Start with the 1. 1 = 6⁰, so 6³ˣ⁺⁵ = 6⁰ equates to 3x + 5 = 0, and x = -5/3.

@user-gr5tx6rd4h

3 ай бұрын

Yes, I saw this in a couple of seconds and it looks quite ridiculous to use logarithms here, except if the point was just training logarithms.

@BluesChoker01

3 ай бұрын

Well, Einstein said it well: "Always describe a problem as simply as possible, but no simpler." Same for programming, don't reinvent the wheel--unless you discover a better wheel."

@DawnKekana

3 ай бұрын

@@BluesChoker01do you have any idea on where to start with programming…I tried to instal the python app in my laptop but it declined now I don’t know what to do from here ..

@girdharilalverma6452

3 ай бұрын

Yes the simplest and logically correct।

@boxvism

2 ай бұрын

That's also how I solved it. But I guess he wants to teach the general method.

Don't need to use logarithms... formally, at least. You just need to remember the zero exponent rule. Just make the bases the same. By the zero exponent rule, 1 can be rewritten as 6^0. 6^(3x+5) = 6^0 => since the bases are equal, then the exponents must be equal. 3x+5 = 0 => x=-5/3

I said x=-(5/3) within 3 seconds. It’s very straightforward when you know n^0=1 for finite n≠0.

@paullambert8701

4 ай бұрын

Yes, I think this a very roundabout way of doing this. He should remember that math tests have time limits.

@johnwythe1409

4 ай бұрын

@@paullambert8701 Definitely the quickest way when we know n^0 = 1, but if it is not =1 how do you solve?

@terry_willis

4 ай бұрын

@@johnwythe1409 :Yes, this only works if YadaYada =1. If not, then :(((

@baselinesweb

3 ай бұрын

Well aren't you just the smartest boy?

@WitchidWitchid

2 ай бұрын

@@paullambert8701 If you use logarithms it is also fairly quick. Reason it took so long in the video is because he spent several minutes teaching what logarithms are and how to use them.

Fast take - this is a thing-to-zero-power because the result is 1. That means whatever it takes to get 6 to the 0 power. Thus 0=3x+5... ergo -5=3x so x=-3/5

@AllDogsAreGoodDogs

5 ай бұрын

Thank you!

@olivemd

5 ай бұрын

How I did it too.

@AquaporinA1

4 ай бұрын

Your fast take was wrong. You put x=-3/5; the correct answer was x=-5/3. So the moral to the story is to slow down.

@siamaknormani2733

3 ай бұрын

Precisely,,

@dumitrudraghia5289

3 ай бұрын

ERROR FINAL!

Greetings. The answer is negative 5/3. The expression 6^(3X+5)=1 can be rewritten as 6^(3X+5)=6^0. Therefore, 3X+5=0 by equating the values of the exponents. Now, finding the value for X is simple. We will transpose the known value 5 across the sign of equality to get 3X=-5, and X=-5/3 after dividing both sides by 3. Lovely.

@andrewhines1054

6 ай бұрын

Even less work than my solution - good job!

@devonwilson5776

6 ай бұрын

@@andrewhines1054 Greetings. Blessings.

@ussdfiant

6 ай бұрын

That’s how I did it.

@williamniver6063

6 ай бұрын

Good, correct, and detailed answer, Devon. I am a down, dirty, and done type of guy: (in my head) "Anything raised to the 0 power equals 1, so 3x+5 = 0, x=-5/3, next?) If Math Man John wanted to illustrate how to manipulate exponents and logarithms to solve the more generalized problems as he does in his explanation, he should have used a different problem. By using "1" here, he invites the analysis you and I did, but if it had read "... = 28", the "exponent is 0" trick wouldn't be useful.

@jim2376

6 ай бұрын

👍

I'm 72, previously an engineering student, and sea captain. Did the sea captain thing after enginering calculus without a calculator got the best of me. Only to be confronted with the joys of spherical trigonometry that is required to do celestial navigation. I'm just saying you younguns with your GPSs and calculators don't know how easy you've got it. I watch your program to help me from forgeting everything I ever learned.

@oahuhawaii2141

2 ай бұрын

But the Earth isn't a perfect sphere. There'll be errors in navigation that needs adjustments to follow the appropriate path to the next port. Otherwise, you could encounter something unexpected and bad.

@WitchidWitchid

2 ай бұрын

@@oahuhawaii2141 Which is what they do. Since it's very close the spherical trig gets you quite close to where you need to be and then you can make the adjustments that is part of the navigation process.

@Kualinar

11 күн бұрын

If you need to navigate at sea, you still need to be able to find your location the old way. Your GPS may be broken or fell overboard.

I like the detail you went into here. I think people who want, or need, context will appreciate it, too. Everyone else probably just wants a quick solution or trick. No dig at them, they probably know most of it. I'm just saying not to be discouraged by people who want you to rush. Videos like this help many who need it.

6^(3x+5)= 6^0 3x+5=0 x=-5/3

The log of a number is the *_power_* to which a *_base_* must be raised to give the number. You can iinvent your own base, but using the example given, log (base 2) of 16 = 4, because 4 is the power to which the base (2) must be raised to give 16.

So it would seem that many of you are brilliant students who were able to recognize that any number to the zero power is equal to 1 and were able to solve this problem in record time... bravo. What you missed was the brilliance of the teacher to use this problem to be solved using the rules of logarithms, that many may not know or possibly are rusty with, and through his steps have given the student an opportunity to see that these crazy rules actually work and the result actually comes out to a value that we can accept and know to be true. The actual student now has validation of the methods that have been shown result in an answer that they can validate.

generally for equations a^(bx + c) = d, you can take ln of both sides so ln(a^(bx+c)) = ln(d), then use the logarithm power rule to bring bx + c to the outside, yielding, (bx + c)(ln(a)) = ln(d), divide both sides by ln(a), bx + c = ln(a) / ln(d), by logarithm rules ln(a) / ln(d) = log base d of a which i will write as log_d(a), subtract by c to get bx = log_d(a) - c, divide by b for the final answer of x = (log_d(a) - c) / b. if we plug in the numbers, a = 6, b = 3, c = 5, d = 1, x = (log_1(6) - 5) / 3, log base 1 of all positive real numbers (including 6) is 0, so x = -5/3

You helped me remember that I know how to do these. Thank you.

Great work, Jon. Can I get into pre calculas with you ?

Good information and review.

I learned logs back in the day but I didn't even go that long route to the problem, because I also learned that any number to the zeroth power equals 1, so I solved this in my head by just thinking "3x-5=0", and then solving for x - simple math, without using a calculator or log table. I don't care what the base number is.

Another method would be to take the log of both sides and one would get (3x + 5)log6 = log1. One would then get 3x + 5 = log1/log6. The next step is 3x = log1/log6 - 5. The solution would be x = (log1/log6 - 5)/3.

set the exponent expression = 0 and solve for x --> 3x + 5 = 0 b/c base^0 = 1 3x + 5 = 0 3x - 5 = -5 x = -5/3

How many used Log tables in the back of your Algebra 2 book and use Trig tables??

Calculators came in the mid 70s, and they were very expensive. And you are a very good teacher. Thanks.

@malenor4148

3 ай бұрын

I took apart my 1970s Texas Instruments calculator and activated the square root button.

@oahuhawaii2141

2 ай бұрын

My sister got a TI SR10. It has built-in [x²], [√x], and [1/x] functions, but no parentheses or memory store/recall. It can display the result in mantissa with exponent format. Neatest thing since sliced bread for engineers and scientists.

Its a easy solution Take 1 = 6^0 bcoz anything power to 0 is 1 Therefore 3x+5 =0 X= -5/3 No need for this long process

I now know where to start, thanks to your relentless teaching and a lot of practice. Exponential equation? Think Logs! It worked. :)

@WitchidWitchid

2 ай бұрын

Anytime you see a more complicated equation or expression in an exponent try using logs to simplify it.

@aryusure1943

2 ай бұрын

@@WitchidWitchid Indeed! :)

6^(something) = 1 mean that (something) MUST equal zero. So, 3x+5 = 0. This changes that exponential equation into a simple linear equation. 3x = -5 → x = -5/3 In this specific case, you don't need logarithm. I do remember using those logarithm tables... It was quite tedious.

What application do you use for your presentation. What application do you write abd draw on please?

I was in my final two years of High School (5th and 6th Forms or Years 11 and 12) here in Australia when scientific calculators became available, but they were equivalent to about a week's pay or more so not many people could afford them. Also, we were not allowed to use any calculator in exams. Calculators were only used to double check our answers and even then, you could not trust calculators to be correct at anything other than basic addition and subtraction.

@devonwilson5776

6 ай бұрын

Greetings. I can remember those days very well.

@davidhandyman7571

6 ай бұрын

@@devonwilson5776 I also remember using a slide rule and it cost a whole lot less than even a basic calculator.

@roger7341

4 ай бұрын

I was a poor graduate student armed only with a slide rule when scientific calculators first became available. I was the only student not using a scientific calculator because I also had a wife and kids to feed. Scientific calculators should never be banned from any exam when they are not banned from the job. Instead, the student should be required to justify every step taken to arrive at any answer obtained from a scientific calculator. For example, suppose my boss handed me this new equation he just discovered, (2.3)^(2x)-4*(2.3)^x+4=0, and ordered me to have a value for x by tomorrow. And by the way, scientific calculators are not allowed. I know several ways to solve this problem, but I can't think of a useful way to get a value for x without a scientific calculator. For example, substitute y=(2.3)^x into the given equation to get y^2-4y+4=0, which has roots y=[4±√(16-16)]/2=2, so (2.3)^x=2. Of course I could then take the natural log of this equation to get x*ln(2.3)=ln(2) and solve for x=ln(2)/ln(2.3), but without a scientific calculator I couldn't provide a numerical value for x without consulting my old "CRC Standard Mathematical Tables and Formulae" book.

@oahuhawaii2141

2 ай бұрын

If you can't trust those calculators, then they're useless. What you couldn't trust is the user, if he/she didn't understand how the device operates and its limitations. I think the real problems are that most early ones had no support for extended precision, floating point (mantissa & exponent), precedence rules, parentheses, and memory store/recall. They also lacked many common functions, such as roots, powers, exponentiation, logarithms, circular & hyperbolic functions along with their inverses. That means you still need to make sure you don't exceed the range of the calculator, use look-up tables for the common functions, be wary of precision errors introduced, and have scratch paper on hand to record intermediate results that will be hand-entered later on. That's where errors get introduced into the calculations. My sister got a TI SR10. It has built-in [x²], [√x], and [1/x] functions, but no parentheses or memory store/recall. It can display the result in mantissa with exponent format. Neatest thing since sliced bread for engineers and scientists.

I like your quick crash courses.

Yes, log tables and all the other tables were extremely tedious to work with. Calculators are a blessing. As is KZread and computers and smart phones. We forget how lucky we are these days.

@paullambert8701

4 ай бұрын

I still like slide rules.

Sir you are underrated you deserve more views and subs

@pauldalnoky6055

6 ай бұрын

Views is what count$

6^(3x+5) = 1 = 6^⁰ Therefore 3x+5 = 0. Exponent values are equal 3x= -5 X = - 5/3 QED

No need for log. Any number to the zero power is always 1. So, 3x+5=0. 3x is -5. Divide bot side by 3. X equals -5

Good info!

Hi John you are an excellent teacher. I am having a lot stuff cleared up

Went to school from 1972 through to college leaving in 1985 at eighteen and parents bought me a Casio scientific calculator in 1981 for my Maths O levels and Physics.

Hi. Great to see this. I must have done it at school but I don't remember it. As for pocket calculators they were not available until maybe 1970 when Sinclair started selling a pocket calculator at an affordable price, (in the UK) this calculator did only basic arithmetic, but very soon Texas instruments brought out a significant scientific programable calculator, a little too large to fit a shirt pocket it had most of the features you might expect today. I bought one to use at work in about 1972/3, it meant calculations that previously needed seven figure logarithm tables and took hours now took minutes. At the time I was working as a design draftsman and still an apprentice. The calculator literally cost a weeks pay. That calculator lasted until the early 1990's when I had the chance to do an engineering degree, I bought a new one then, it cost about the same in money as the previous one. I see today a similar Texas Instruments calculator costs much the same, which in real terms is many times cheaper.

@oahuhawaii2141

2 ай бұрын

A friend got a Sinclair calculator. It was fun to play with, but we kept draining the batteries. He also got the computer, too.

@oahuhawaii2141

2 ай бұрын

My sister got a TI SR10, and later my brother got the HP 34C. I got a TI 35, TI 57, and HP 41C. Another friend got a TI 58. Another student got the TI 59. They were fun to use.

Simpler: If (3x + 5)log 6 = log 1 = 0, then since log 6 not= 0, 3x + 5 = 0, and x ≈ -5/3.

I got the answer by accident. I learn so much from your videos. Thank you!

I am so happy I never had you to teach me math.

@OR161NYT

2 ай бұрын

you funny man, i like u

6^(3x+5)=1 so taking logs of both sides and dealing with the exponent we get (3x+5)*log(6)=log(1), and thus 3x+5=log(1)/log(6), but log(1)=0 (for logarithms of any base) so 3x+5=0, so x=-5/3.

@kennethwright870

5 ай бұрын

For that equation to be true, 3x+5 must equal 0, so x=-5/3

On Indices any number raised to zero gives 1 therefore the power on the right is zero meaning equating 3x+5=0 solving gives X=-5/3 .The explanation may confuse learners.

6 to the power of zero equals 1. Hence, 3x+5=0; this yields x=-5/3

I'll solve this equation in three different ways, and one should know all three of them. (1) 6^(3x+5)=6^(3x)*6^5. Divide both sides by 6^5: 6^(3x)=6^(-5). Equate powers of 6: 3x=-5 so x=-5/3 (2) Any number raised to the zeroth power is 1, so 6^0=6^(3x+5)=1 yields 3x+5=0 and x=-5/3 (3) The natural log of 1 is 0, so take the natural log of both sides: (3x+5)ln(6)=ln(1)=0, so (3x+5)=0/ln(6)=0 yields 3x+5=0 and x=-5/3

Answer: -5/3. Take the natural log of both sides, bring the exponent down on the left side. (3x + 5) ln (6) = ln (1) ln(1) is equal to zero: (3x + 5) ln (6) = 0 Divide both sides by ln (6): 3x + 5 = 0 Subtract both sides by 5: 3x = -5 Divide both sides by 3: x = -5/3

@FormlessFlesh

6 ай бұрын

You can also take the log instead of ln. Either way, you get the same answer

Maybe you should have had this equation equal something other than 1, since knowing that anything to the zero power =1. So if 3x+5 is zero, then it's really easy and doesn't require logs. Anyway, thanks for the instruction on logs. They always kinda perplexed me, and you're helping me to understand.

Good lesson

Take logs: (3x +5) × log 6 = log 1 (note that log 1 = 0) therefore (3x + 5) × log 6 = 0 divide each side by log 6: 3x + 5 = 0 so x = -5/3 = - 1 ⅔

Yes many won’t no where to start because they preoccupied by the equation of how to make there paycheck equal to the cost of living.

Yep, when the bases are identical, they can be eliminated from exponential equations. Of course, this is a *nice* example because "1" can be replaced by 6^0. The change of base formula is unnecessary. 😊 To generalize, if equation were: 6^(3x+4)=36 then 6^(3x+4)=6^6 3x+4=6 3x=2 x=2/3 Chk: 6^(3*(2/3)+4 )=6^6 6^(2+4)=6^6 6^6=6^6 36=36 This identity holds for both real, complex and trigonometric exponents.

@dianegesik7456

2 ай бұрын

Thanks for providing the explanation. It is helpful. I am arriving at a different answer than yours however. Could you take a look to see if you agree? In the third line of your calculation you are showing 6^(3x+4)=6^6. I am asking if this should instead be 6^(3x+4)=6^2. Thus 3x+4=2, 3x=-2, and x=-2/3.

Just start taking logs from both sides, which eliminates the 6, and solve the equation.

At first it looks like a hard and tricky problem due to the x variable in the exponent. But it is really a very easy question.

Don't forget to mention the slide rule!

@BluesChoker01

3 ай бұрын

We used slide rules when I was a junior, but we could use calculators as seniors. The school bought them and we could borrow one. The first Calc I used was the HP which used polish infix notation. It did everything but graph. Switched to TI as undergrad. Returned 20 years later for advanced degree, and we used a nice--TI graphing calculator. But for advanced Calc, discrete math, etc--we used laptops and Maple and Mathematica (now Wolfram). Once mastering these tools, you just couldn't miss homework problems. The best thing about these tools, was the ability to try all sorts of approaches. If an approach failed, you learned something. Playing around and thinking really expands your depth of knowledge. One thing to remember is there are no timed tests in the workforce-- only deliverable dates. Sure, some things needed quickly reviewing, but some problems took months, so they were always in the background, to be attacked as insight lit up connected neurons :-) Why is it that most "aha" moments occur in the shower, working out or in dreams?? 😮 I think the subconscious is handling a lot of work so the conscious mind can function on the immediate tasks at hand. Man, I got pretty metaphysical here.

Is someone able to build a solution in Excel/Google? I'm having a hard time solving for a complex exponent in Google/Excel. This is based on a financial formula. A=P*((1+r/n)^(n*t))+x Solving this for (t) (A-x)/P =(1+r/n)^(n*t) At this point, I think I need to use a log function to get the exponent out, but if I capture (t) in "log(1+r/n,n*t)", I'm really not sure what to do to get (t) out of the function. Please help!

wow, i actually deduced that the exponent had to be 0 for the equation to = 1. dang, sometines i amaze myself. all in fun. thanks for a great lesson, & actual reassurance. i went to HS in the mid-late 60s. if you got caught with a calc, YOU FAILED, sometimes the test or even the semister. DONE !!!

@user-ri6rn7ti5h

6 ай бұрын

All good thing comes in three's

@devonwilson5776

6 ай бұрын

Greetings. I remember those days, for me 70's, in primary school, absolutely no calculator allowed. The good old days.

@ndailorw5079

5 ай бұрын

That’s because calculators are for students who understand and therefore know how to do the mathematical calculations without using a calculator. What if the calculator broke down and we didn’t know how to arrive at the answer way the instructor is doing here? It’s easy to hit 5 and then log10 on the calculator. But what if the calculator pegs out and break right at that moment and we didn’t know how to do the problem by hand because we didn’t know and understand how to do the problem by head and hand? Calculators are excellent for doing tests and even during class, but only after a subject has been thoroughly understood and learned by the head and done by the hand repetitiously until committed to both memory and understanding. Does no good to have students hit and peck 2 + 2 on the calculator when they don’t understand and know how to do addition through understanding and repetitive practice. What happens if the instructor gives a set of problems to be done within a reasonable time frame but says, “absolutely no calculators!” Then what! Calculators are for those who understand and know the work, they’re not for the uninitiated! The uninitiated will never understand and learn the subject matter properly and thoroughly by simply punching in log100 with a calculator… they need to first understand and learn why the calculator gives the answer 2! For example, why would I add 0.3456889731237 + 35.903578134556 = x by head and hand instead of using a calculator when I already understand and know how to add decimals by head and hand? It would consume way to much time and be completely insane for the teacher to have me to write that problem out by head and hand instead of using a calculator when the teacher knows that I’ve proven and therefore understand and know how to do the problem by head and hand! But for the uninitiated student they must necessarily begin by first understanding and knowing and doing the problem of adding decimals before they move on to calculators. Calculators are horrible for students who don’t first understand know how to do math by head and hand through practice, practice, practice! So… what do I do with 30.4567893400586 - 7.8967855422106 if I didn’t know how to subtract decimals by head and hand but knew how to peck out the answer on a calculator but all the calculators in the world were broke….?

Start with the exponent. It has to yield 0 since 6^0 = 1. So 3x + 5 = 0. x = -5/3

Any number to 0 power is 1 3x+5=0 done.

I am nearly 70 years old. There are "gaps" in my math. These are because of poor teaching when I was in school. Just in case you think I am/was the problem, I was always at or near the top of my class in Math and achieved being in the top 10% of the State in my Higher School Certificate. Having someone to actually teach you correctly is essential. By the way, I correctly got the answer in about 15 seconds without using logs.

@rremnar

6 ай бұрын

I can vouch for this. When I went to Lake Stevens HS, the geometry teacher was a total bore. I didn't learn much and I had a hard time paying attention, so I would up with a D grade. When I transferred over to Mariner HS, I got into a trigonometry class. The teacher there was good, and I got a B. The teacher does make a huge difference, but the door swings both ways.

Back in the good old days you used a slide rule. I used tables and a slide rule in 1955.

@oahuhawaii2141

2 ай бұрын

Yes, but in his 2 problems you can do them easily, right? 6^(3*x+5) = 1 Take log base 6 of both sides: 3*x+5 = 0 x = -5/3 ≈ -1.666667 4^y = 10 y*log(4) = log(10) y = 1/2/log(2) We remember log(2) ≈ 0.30103, so y ≈ 1/0.60206 ≈ 1.660964 .

A much easier way. 6 to the zero power is 1. As such 3x + 5 = 1. Took me less than 3 seconds to solve.

Multiply both sides with 6^-5 => 6^3x = 6^-5 => 3x=-5 .............

6 to the zeroth power is equal to one. Easy to solve now, right?

Log tables were only for when you needed more precision than your slide rule could give.

6^(3x+5)=1=6^0 compare indices, 3x+5=0 x= -(5/3). It took me about 3 secs.

Any number raised to the zero power equals 1. So if 6 raised to the power 3× +5=1. It implies that 3× +5= 0 and when you solve this ,it gives × = - 5

@nickcellino1503

3 ай бұрын

No. It's -5/3. You forgot to divide 3x by 3.

I’m not sure how to solve it, but I know the exponent needs to be 0 fir this to make sense. I divided by 3 and got -5/3, but this was an expression, not an equation, and in an expression I can’t extricate the variable from it, as I understand it.

the inverse of exponent is logarithm, Why do i keep thinking is square rooting? Whenever i rearange a formula with exponent, i do sqr rooting on both side as inverse or? Sir can you do a vid on the difference between the two please.

@ndailorw5079

5 ай бұрын

You’re right, when and where it applies. For example, x^3 = 27, which gives x = 3 after taking the cube root on both sides of that equation. Or, the same thing for x^2 = 100, say, which, taking the square root on both sides gives x = +/-10, apart from dimensions of time, length, etc., since there’s no such thing as negative length and time, etc. in which cases only the positive answer and not the negative answer is possible. Or, and probably more to your question and example but the very same thing nonetheless, and simply the same thing the other way around but using the inverse of roots we could have the square root of x = 10 where we’d then square both sides of the equation (the inverse of taking the square root on both sides) to get x = 100! Or we could have the cubed root of x = 3, where we’d simply raise the expressions on both sides of the equation to the third power (cube both sides of the equation, sort of speak) to get x = 27. Well, here, the same thing’s going on since logarithms and exponentials are inverses of each other, like exponents and roots are, one to the other, just as addition is to subtraction and division is to multiplication, and vice versa, etc. That is, we can use the inverse of a mathematical operation that’s in the form of an equation to solve that equation, or to check for the correctness of it. In other words, we check our answers to our addition for their correctness by its inverse operation of subtraction, and vice versa. Same with multiplication and division with respect to each other, etc.

Step #1, convert 1 into 6^0, then with the same base, exponents must be equal.

@WomenCallYouMoid

6 ай бұрын

3x+5 = 0, 3x=-5, x=-5/3.

@WomenCallYouMoid

6 ай бұрын

Especially easy when it's equal to one & only done so when it's equal to one. You can also do logs on both sides & doing a log of 1 is equal to zero & a log base 6 of 6^z gets you z = 0 |z=3x+5 & u get the same thing

@jamesharmon4994

6 ай бұрын

@Mike-lx9qn Yes, but it seems unnecessarily complex and does not teach the principle that when bases are equal, then the exponents must be equal.

@thomasmaughan4798

5 ай бұрын

@@jamesharmon4994 "bases are equal, then the exponents must be equal." However, 5^0 = 1, 6^0 also = 1, so it is not clear that one must choose 6^0. Instead it is probably sufficient to know that any number to the 0 power is 1, thus the exponent must be equal to zero and it isn't really necessary to put the same base on both sides of the equal.

@jamesharmon4994

5 ай бұрын

@thomasmaughan4798 True, but if the instructor requires "Show your work", it is.

For any y, y^0 = 1 Therefore, 3x + 5 = 0 3x = -5 x = -5/3 Where is the challenge? I did this in my head.

Yeah, the ‘70s……….we had to actually DO this lol! Ah, the tables in the back of the books. We were allowed to use the NEW scientific calculators in my senior year (1978). They were exorbitantly priced (over $100)! The minimum wage was $2.65 then. It would be like paying over $275 today. My dad was unhappy! I had a friend who used his trusty slide rule. Ah, those were the days!

Take log base 6 of both sides to get 3*x+5 = 0. Then, we see x = -5/3 .

@oahuhawaii2141

2 ай бұрын

If you want to go for the infinite solution route with complex numbers, then you need to consider the polar form for 1, which is 1*e^(i*2*π*k) for all integer k: 6^(3*x+5) = 1 = 1*e^(i*2*π*k) (3*x+5)*ln(6) = 0 + i*2*π*k x = -5/3 + i*k*(π*2/3/ln(6)), for all integer k Note that for k = 0, x = -5/3 .

A^0=1 when bases are same powers are equated so 3x+5= 0 , so x=-5/3

Most simple way is to make the power of 6 zero, so because 3x +5 to be zero needs 3x to be equal to -5 so x should be -5/3

It's easy, exponential and logarithmic

It is Not the Answer that you are learning here. Its the Method so can apply to every others. In Maths Methods solves.

There’s another way and is elevate both terms of the equation to 0 And so forth and so on.

This can be solved using derivatives, right?

6^3x+5= 1 6^3x/6^5=1 6^3x =6^-5 base 6 is the same on both sides 3x = -5 X =(-5/3) For checking this value in the above equation 6^3(-5/3) +6= 1 6^(-5 ) +5= 1 6^0= 1 1 =1

@DawnKekana

3 ай бұрын

What happend to the 1 when you moved to step 3?

If we assume that any number that has 0 as an exponent is equal to 1, then 3x+ 5 = 0

4^=10= 3 .2 (x+2x-3)

Any number not zero, to the power of zero equal 1. Thus 3x+5 = 0, or x = -5/3.

please say right lets dive into it so it sounds more professionial and to the point period and for also how it should sound instead of just t alking which sort of isn't really interesting oj its own before the go to statements ,... ,,,.,,.

any number raised to the power of 0 is 1, therefore 3x +5=0 and QED x=-5/3

@oahuhawaii2141

2 ай бұрын

Some mathematicians will squawk about 0^0 .

could you use x = -(5/3) to solve the problem?🤔

At the beginning would be excellent

The full solution should be -3/5 + 2*n*pi*i , where is is square root of (-1). n is any integer

@oahuhawaii2141

2 ай бұрын

Nope. You are way off.

@oahuhawaii2141

2 ай бұрын

If you want to go for the infinite solution route with complex numbers, then you need to consider the polar form for 1, which is 1*e^(i*2*π*k) for all integer k: 6^(3*x+5) = 1 = 1*e^(i*2*π*k) (3*x+5)*ln(6) = 0 + i*2*π*k x = -5/3 + i*k*(π*2/3/ln(6)), for all integer k Note that for k = 0, x = -5/3 .

Natural logs of both sides, John?

@richardl6751

6 ай бұрын

He is using common logs, base 10.

@N269

4 ай бұрын

Makes no difference.

Surely 6^0=1, so with a common base of 6 we have 3^x+5=0, ie x= -5/3

So, how is this applied in the real world?

if u use the a^0=1 then 3x+5=0--> 3x=-5-->x=-5/3 isn't it easier?

1 = x^0 , therefore 6^(3x+5% = 6^0, therefore 3x+5=0 x=5/3

I got -1.66 why is that wrong? Should I have left my answer in a fraction? I know it means nothing, but, this is fun. I've had much higher calculus than any of this, but I apparently have a lot to learn?

I would just set 3x+5 as t and solve t first and then substitute it.

Any number to the zero power is 1. Set 3x+5 =0 and solve for x. No need for logarithms or any of that for this problem.

Change 1 with 6^0

i entered log10/log4 and my calculator showed 1.2203,,, as answer not 1.66. whats up? maybe i entered wrong syntax?

@ndailorw5079

5 ай бұрын

Calculators aren’t all programmed the same. Some accept entries one way while others do it differently. Find out what type of hierarchical program your calculator has, or look at its usage instructions and you’ll get the correct answer.

Start with bringing the exponent down

I typed into my calculator 'log10÷log4 and the answer was 1.22 weekday am i doing wrong?

@vespa2860

6 ай бұрын

Using log to base 10 ( the normal log button): log 10 = 1 log 4 = 0.6 approx gives 1.66 when divided Using natural log (the ln button): log 10 = 2.3 approx log 4 = 1.386 approx again gives 1.66 I've no idea how you arrived at 1.22 (even tried mixing the two log functions). edit - I used your 1.22 and swapped around to find either: log 10/log 6.6 = 1.22 or log 5.426/log 4 = 1.22 Doesn't help much! On the Windows calculator (scientific mode) Simply input 10 press log - then divide by 4 press log i.e. (10 log)/(4 log) Not sure what calculator you are using.

@ndailorw5079

5 ай бұрын

…may depend on the hierarchical structure (program) built into your calculator. On mine I hit 10 (the 1 and 0 buttons), then hit the log 10 button, then hit the divide button, then hit the 4 button, then hit the log10 button, then hit the = button and the calculator gives the approximation 1.66……….. how you enter those terms in the proper sequence and correctly on a calculator depends on the calculator’s hierarchy structure.

But you didn't show why anyting raised to zero is one; i.e. why x^0 =1 - I will show it in two ways, the first with numbers to help get the feeling for it, and the other with a in order to "make it mathematically correct as a proof" 16=2^4, 8=2^3, 4=2^2, 2=2^1, 1=2^0, 1/2=2^-1, 1/4=2^-2, 1/8=2^-3, 1/16=2^-4 Beginning with 16, every time you divide by two, the result is that one is subtracted from the exponent, and so your exponent starts at 4 and is reduced to 3, 2, 1, 0, -1, -2 -3, -4 while the number is halved down to 1 and then to 1.2 to 1/4 etc, and the 1 occurs when the exponent is 0. axaxaxa=a^4, axaxa=a^3. axa=a^2, a=a^1, a/a=a^0, 1/a=a^-1, 1/(axa)=a^-2, 1/(axaxa)=a^-3, 1/(axaxaxa)=a^-4. Take a4/a =a3 (axaxaxa)/a =(axaxa). Similarly (a^3)/a=a^2. Similarly (a^2)/a=a^1. Similarly (a^1)/a=a^0. Similarly (a^-1)/a=a^-2. The crucial point is (a^1)/a=a^0. i.e. a^1 = a, and a/a=1 while a^1/a^1 = a^1 x a^-1 = a^(1-1) = a^0. So we have a/a = a^0 and a/a=1

I started with x = -5/3 Done

Parece fácil : 1=6^0 assim 3X+5=0 X=-5/3 e pronto !!! 🤷🇧🇷

I think there is something wrong with n ^ 0 = 1. I'll explain. When you write out a exponential term in multiplication, it includes the original number itself. So n ^ 3 = n * n * n . If you have n ^ 1 , then the answer is n. You got that? What happens when the exponent is 0? Then we have no numbers to multiply with, not even the base number itself; so the answer should be 0. Now I researched this and saw why it was 1, but it is still incorrect. When you have a exponent, it not only needs to be a positive real number, it has to be greater than 0, for the base to exist. Edit: The reason why they say n ^ 0 = 1 is because n ^ 1 / n ^ 1 = 1 which could be restated as n ^ (1 - 1) = 1, so n ^ 0 = 1; but as I explained above, that can't be right. So what happened?

@ndailorw5079

5 ай бұрын

First, If, as you say, when the exponent is 0 we have no numbers to multiply with, how then did or would we get or assume we’d get 0? Wouldn’t the answer as well as the question then be meaningless, if not absurd? And why do you say an exponent needs to be a positive real number? -3 is a negative real number and therefore less than 0 and bases have negative exponents such as -3. So are you saying that the base x in the expression x^(-3) doesn’t exist because negative 3 is negative and not positive and less than 0? Are you saying that we can’t have a negative exponent such x^(-3)! What about x/xxxx = 1/xxx = 1/x^3 = x^(-3)? That’s the same negative exponent, isn’t it, and it’s base exists! If the rule of division for exponents states that x^a/x^b = x^(a - b), then If a = b and we have x^a/x^b, then x^a/x^b = x^a/x^a, since a = b, = 1 =, by the rule, x^(a - a) = x^0 = 1. The proof (x^n/x^n = 1 = x^(n - n) = x^0 = 1) justifies the rule. Which is the same thing as x^a/x^b = x^(a - b) x^2/x^3 = x^(2 - 3) = x^(-1) = xx/xxx = 1/x = x^(-3). So that, where (a) equals (b), for example, both (a) and (b) equal 2, then, x^2/x^2 = xx/xx = 1 = x^(2 - 2) = x^0 = 1. You’re simply subtracting the power of the denominator from the power of the numerator. Which simply says that when the two powers (exponents) and their signs are equal and the same the quotient has to equal 1. By implication the rule is simply a logical and true conclusion, is all.

more known as Shiva unless misaken not the actual god the Ramanga mathmatician known as Shive as welll where it is Ramanag Shiva but i feel a bit wrong where it was and am will to take critisome on the 2nd name

How about this 6^3x+5=6^0, then 3x+5=0 and x=-5/3... Fast and easy