Here's a video with 10 examples of solving exponential equations, from basic to hard!: kzread.info/dash/bejne/fWx3s8SMdJzdmdY.html

I'm just impressed how you write with two different colors in one hand.

@dddaaa6965

3 ай бұрын

Imagine now how sex

@loser1233

3 ай бұрын

it's like using chopsticks

@yafmaverick

3 ай бұрын

You must be special

@viCuber

3 ай бұрын

Lmao exactly the same I commented about two weeks ago

@kajalde3071

2 ай бұрын

I'm impressed that you do that with two hands

I don't know why I'm watching this at 5AM since I'm a physicist doing PhD in neurophysics and computational neuroscience, but I thoroughly enjoyed this. 10/10. Younger generations are so lucky that they have someone like you explaining maths. Hopefully they'll know how to appreciate it and not waste their brains away on TikTok...

@scienceislove2014

2 ай бұрын

Woah..sounds interesting.. can you elaborate like what things you study and tools you use?

@johnmarcusengreso8273

2 ай бұрын

Ill take it as a compliment mr neurophysics man Im a 10th grader that likes math

@J0EB0B555

2 ай бұрын

I'm a physics major as well but haven't taken a math course in a while. He's really helpful for keeping all the concepts fresh in my brain.

@IskzenMisishuw

2 ай бұрын

Yes brother true

@antonioruelas8902

Ай бұрын

Physics undergrad here, this man (and Organic Chemistry tutor) saved me during calc 2

"How to find X?" Bro, it's time to move on. Your X doesn't care about you anymore.

@bprpmathbasics

2 ай бұрын

y?

@TheOnlineTurtle

2 ай бұрын

@@bprpmathbasics😂 thats good

@r0N1n_SD

Ай бұрын

Lol. Bro got owned by Math😂

@stalincomrade1867

Ай бұрын

Don't you z how pointless it is?

@Taokyle

Ай бұрын

lol

i took precalc 4 years ago and was arbitrarily recommended this video yet I still feel compelled to do the homework this man has given

@xxxBradTxxx

Ай бұрын

I took it 15 years ago and I still watch these videos because it feels like a waste of effort to learn all of that an forget it. 🤷‍♂️

I'm in 10th grade, so whenever he says "let's use this rule" I'm just like "uh huh" Edit: it's crazy how different some curriculums are in other countries.

@diewand5442

3 ай бұрын

We learned the logarithm in 10th grade😅 (Germany)

@AutoFun_

3 ай бұрын

Why are u here?

@Gamert80

3 ай бұрын

I enjoy watching advanced math, even if I don't understand it fully.

@Gamert80

3 ай бұрын

@@diewand5442 I'm only half way through the year so I may learn it soon.

@aaryan8104

3 ай бұрын

SAME FROM INDIA BTW

To expand it, you use change of base to get log(96)/log(2/3). When dividing a logarithm, of course, you subtract the log of the denominator from the log of the numerator, which gives log(96)/(log2-log3). We can take the prime factors of 96: 3 and 2⁵, to get log(3•2⁵)/(log2-log3). With multiplication of logarithms, you add the logs of the multiplicands, so (log3 + log(2⁵))/(log2-log3). Finally, with exponentiation, you multiply the logarithm of the base by the exponent, which gives (log3 + 5log2)/(log2-log3).

@jordananderson2728

3 ай бұрын

I should have used ln rather than log, but I'm so used to using log for change of base that I just did that by default. It works the same either way (:

@GreggRomaine

3 ай бұрын

Nicely done, thanks for doing my homework!

@paratoxicalcapybara3939

3 ай бұрын

you can just use laws off exponents bcuz that seems easier. then you log it at the end for answer turn 2^(x-5) into 2^x*2^-5 and turn 3^(x+1) into 3^x*3. an example of this is (3^2)*(3^2)=3^4 expand into 1/32(2^x)=3(3^x) do some division to isolate x as much as possible. 3/(1/32) = 96 or (1/32) = 1/96. End up with 2^x=96(3^x) or 1/96(2^x)=3^x x root everything. 2=xroot(96)*3 or xroot(1/96)*2=3 more division to isolate x. 2/3=xroot(96) or 3/2=xroot(1/96) put everything to the x power. (2/3)^x=96 or (3/2)^x=1/96 now log bcuz inverse of exponential to finnaly actually isolate x. log(base(2/3)) of 96 = x or log(base(3/2)) of 1/96 = x x= ~-11.25

@Kasakuja

3 ай бұрын

well, how do you get from log(96) ---- log(2/3) to ln(96) --- ln(2/3) ?

@TheWatch1

3 ай бұрын

Log(96)/Log(e) is ln 96. Divide by log e in Nr and Dr

That was a really good explanation! Thank you for explaining so clearly! 👏

the second option is always what comes to my mind first, i find it way easier and more intuitive, but ive forced myself doing the natural base method too cuz you have to know them both imo

@riccardodellorto4267

3 ай бұрын

I do the complete opposite: whenever I see an x as an exponent, I use ln, because the calculator can eventually solve any monstrosity I type in as long as there are numbers 😂 Bringing down the x is my number one priority 🫡

So I'm a 3rd year medical student watching this video and I dearly enjoyed it. Its like going down the memory lane. Really smooth teaching. Kudos to you..❤

I have a Zoology exam tomorrow. It's 3am. 10/10

@ominious7082

19 күн бұрын

Yo how did it go? 😂

@utkarshjain861

2 күн бұрын

So, how did it went?

I solved it slightly different. I recognized that 3^(x+1) can be rewritten as [(1.5)(2)]^(x+1), which can be expanded as 1.5^(x+1) 2^(x+1). This is very helpful as it gives us an exponential of base 2 on both sides of the equation, which allows us to cancel out the x on the left side through exponent division rule. The full solution is below: 2^(x-5) = 3^(x+1) 2^(x-5) = [(1.5)(2)]^(x+1) 2^(x-5) = 1.5^(x+1) 2^(x+1) 2^(x-5)/2^(x+1) = 1.5^(x+1) 2^(-6) = 1.5^(x+1) Now we only have a single x variable to deal with, so we could simply apply log to both sides and isolate for x log[2^(-6)] = log[1.5^(x+1)] (log[2^(-6)]/log[1.5]) - 1 = x -11.257 = x

@user-lb3ex6yh9u

3 ай бұрын

Well done

@praneel1059

3 ай бұрын

when i saw the thumbnail i guessed that since 2^(x-5) = 3^(x+5) we can do something like 2^(x-5) . 1/ 3^(x+5) then 2^(x-5) x (3^(x+5)) ^(-1) and go on i guess . Btw im in ninth grade so i have no clue about what ln is

@hridayjr6580

2 ай бұрын

nice is the antilog required or this is it.

@scorpio9711

2 ай бұрын

ln is called natural log, where the base is 'e' which is called eular constant. BTW which country do you belong to

@kashi2928

2 ай бұрын

I got that too, thanks for making me feel like I wasn't alone 😂

I gave up on maths nearly 7 years ago in school. In my post graduation i watch this and feel my antipathy towards the subject reduce a little. Thanks

Thank you for taking the time to make this video. Much appreciated. ❤

@bprpmathbasics

Ай бұрын

Glad to help! 😃

I Just Saw the Thumbnail And Thought " Ehhhh That looks Ez Lets Just Do It " Only to waste 30 mins And Find Out It Have Logarithm Which I Havent Studied😂

to think that I knew all those formulas you used and wrote on right side but still I didn't knew how putting them together will get me the answer. Thanks a lot. Any advice on how I can solve this thing of not knowing when to put and which things together to solve questions like this ?

bro fumbels my brain and proceedes to say:"but, here is a prettier way to do it"

@CST1992

2 ай бұрын

C'mon dude, if you know the rules of log this is a pretty simple problem.

@frostcrackle2374

Ай бұрын

So maybe they don't know logarithm rules yet. C'mon dude if you can calculate a Hohmann Transfer, this is a pretty simple problem. ​@@CST1992

@celoreads

Ай бұрын

@@CST1992 if we know the rules of log we wouldnt be here for an explanation now would we? lmfao

@CST1992

Ай бұрын

@@celoreads you don't know what log is but you are on a calculus video? Go back to high school... "lmfao"

@abcdqwerty3562

29 күн бұрын

@@CST1992Do you not realise that the title of the video literally says precalculus?

You are so talented in teaching. Thank you for your wonderful videos.

Very first approach of solving exponential equations is using logarithms.

australian here, i used my calculator. ive only seen the thumbnail and came straight here. the answer i got was (-ln(96))/ln(3/2) or approximately-11.257 edit: finished the video now and checked those two values of x. both were equal to my above answer. very nice 👍

Bro I'm in 10th grade and I reached (2/3)^x = 96 and was like, "Now what?". Then I realised "Oh, this is out of bounds" 💀💀💀💀

@imagod4796

3 ай бұрын

this is 9th grade in Germany

@ryg4493

3 ай бұрын

@@imagod4796 I thought Asia had the toughest math.....

@exip9288

3 ай бұрын

@@imagod4796 This is 12th grade in Turkiye (I know it sucks dumb education system) , but I learned it way before because of calc bc.

@bruv4266

3 ай бұрын

@@exip9288 We all have shitty educations, here in Romania we learn calculus in 11th grade to 12th grade, they should have system of education like USA, this is where the people can learn it well, we have short time in school but too much to learn, cause it's not just math, its also other lesson that it supposed to be in college like physics, chemistry, etc.

@NotKartikeySingh

3 ай бұрын

bro but if u have studied from better school in 9th they would have taught u (in india)

I would consider simplify it with log to the base 10 which yields the same answer as the answer you obtained. We could write it as, X-5log(2)=X+1log(3) Which on further simplification can provide, x= -6.58/0.58= -11.3 And the answer you obtained at the end, log (base)2/3 (96)= -11.26 (approx) I feel its less hectic

I'm not sure which form of the result I prefer. The first form uses ln rather than a logarithm with an awkward base, but the second one looks neater.

I have a question sir. Why we need to use ln instead of log, or we can use which?

I did it the first way, but with logbase2 instead of ln. Thanks for the video!

Could you please make a video on how to find values with decimal exponents Example) (15) ^1.4

Love this reddit series

Another way to do it like the first method that isn't exactly any faster but came to me is: once we have (x - 5) ln(2) = (x + 1) ln(3), (eq. 1) we can build h(x)=(x - 5)/(x + 1) = ln(3)/ln(2), (eq. 2) which will have the same solution despite the domain changing a bit, since the solution isn't near -1. and separate that into two equations: f(x₁)=(x₁ - 5) = ln(3) g(x₂)=(x₂ + 1) = ln(2) so solution x (to h) will be formed by solutions x₁/x₂ to f,g respectively. Which, is a linear system. now we can produce a matrix: [[1, -5, ln(3)], [1, 1, ln(2)]] which we partially row reduce to [[6, 0, 5ln(2) + ln(3)], [0, 6, ln(2) - ln(3)]] recombining, since solution to h is solution to f over solution to g, the 6's cancel and we have: x = (5ln(2) + ln(3))/(ln(2) - ln(3)) which is also the solution for eq.1 I know written out this seems long, but it went a lot faster in my head. A lot of the steps here would be incorrect if I didn't explain them carefully. Also it would probably take longer to row reduce than just doing the algebra but ¯\_(ツ)_/¯ let me know if I did anything illegal math manipulations

@argonwheatbelly637

2 ай бұрын

I like it. 😊 It reminds me of when I use Synthetic Division to blitz through some polynomial division by hand.

Immediate reaction is "x is not positive integer, because 2 and 3 are prime, so the prime factorisation of 2^i will never equal that of 3^j, where i and j are any positive integer".

@tobybartels8426

3 ай бұрын

This also works for negative integers, even for non-zero rational numbers. So the only possible rational solution would be if both exponents are zero (at the same time, which is not possible in this case).

@deltalima6703

3 ай бұрын

You never know if x is a quaternion or is mod |p| or whatever in these dumb questions.

@General12th

3 ай бұрын

​@@deltalima6703This is an algebra channel, not a calculus or analysis channel, so don't overthink it.

@jackposiedonforever7774

3 ай бұрын

Correct

I learned this a while ago, i kinda just forgot about it. So yeah, im greatful for the recap

Thank you sir ❤

Try this next: 2^x=5^(x+2) Answer here: kzread.info/dash/bejne/iYBh0NKMddy8hNI.html

@imabiggoofy

2 ай бұрын

bro really out here assigning hw 💀 (I'm in 8th grade, i dont know shit)

@FloraLemonYT

2 ай бұрын

I actually learned this last unit. 2^x=5^(x+2) xln2=xln5+2ln5 xln2-xln5=2ln5 factor out x x(ln2-ln5)=2ln5 x=2ln5/(ln2-ln5) I’m not sure if there’s a better way to simplify it

@IamFlaem1

2 ай бұрын

x=log2/5(25)

@Neet_mbbs.0907

Ай бұрын

​@@FloraLemonYTThat's correct! 👍

@vandernight1220

Ай бұрын

I mean: (X-5)log2 = (x+1)log3 … -> x = (5log2 - log3)/(log2 - log3) is just way less complicated than the methods shown, at least this is the standard method in uk

Actually the equation becomes easy, when you use log in exponential problems. Thanks ❤🇮🇳

@slulzspot7583

3 ай бұрын

मुझे भी equation देख के वही लगा।

Hey there! Could I just use Log10 instead of Ln, so that the first equation would be Log(2^(x-5)) = Log(3^(x+1))? It comes more naturally to me this way and, knowing that it's added to the two sides of the equation, seems like both ways would be accepted. Is that right? Thanks!

@edwardSteadyHands

2 ай бұрын

Yes doesn't make any diff which base you take

I wasn't taught log at school at all. I had to look it up online. Even though we hadn't had proper knowledge about log we still have to use in calculus

@Roro-ej7ke

Ай бұрын

Lol same💀

Bro i am a Engineering major why did i click on this video

@actionj761

14 күн бұрын

Same thing im like do i really have nothing better to do than to glance at my freshman year history 😂😂

I solved it in a similar way somewhat. Started with taking the natural log but instead grouped terms like: (x-5)/(x+1)=ln3/ln2 (x+1-6)/(x+1) = ln3/ln2 1-6/(x+1)=ln3/ln2 (x+1)=-6/(ln3/ln2-1) x=-6/(ln3/ln2-1)-1 x~=-11.257

@nothingbutpain863

Ай бұрын

That seems more intricate.

you can do backwords in 6:00 ONLY IF a and b are both positive (theoretically a can be 0 , but it's a disputable question)

@amanda-we9fv

3 ай бұрын

что?

@lukaskamin755

3 ай бұрын

@@amanda-we9fv now it's correct. I mean you can't do backwards if a and b are both negative ,roots of a and b won't be defined then, while root of ab will be defined

@arcturusgd

2 ай бұрын

In this case it is. If it is positive that means a,b ∈ N

Wow that is much easier than i thought

It’s 4:20 am right now and I have no idea why I’m watching this at this time. I told mom to call me at 8 and wake me up. I guess now I have a solid reason to tell her why i was awake.

guys we can solve it in another way too. what i did was this: i took log on both lhs and rhs. so the exponent comes down and the equation becomes like such (x-5)log 2=(x+1) log 3 now we know log 2= 0.3010 and log 3=0.477 so we just use those values in the equation (x-5)*0.3010=(x+1)*0.477 0.3010x-1.505=0.477x+0.477 this becomes -0.176x=1.982 x=1.982/-0.176 x=-11.26

@user-bg9xo3ub8q

2 ай бұрын

It was not log tho. It's was ln.

@sahhanaaa

2 ай бұрын

@@user-bg9xo3ub8q works w log too

@Gaysandthechaos

2 ай бұрын

​@@user-bg9xo3ub8qyeah you'd have to multiply it with 2.303 to convert ln to log That'd be easier ig

@user-bg9xo3ub8q

2 ай бұрын

@@Gaysandthechaos yes

@nothingbutpain863

Ай бұрын

​@@user-bg9xo3ub8q, in this scenario, either 'ln' or 'log' is acceptable. This is because the bases of logarithms would get cancelled in the process as long as the bases are the same.

In the first example, could you have used log instead of ln? When to use log vs ln?

@nothingbutpain863

Ай бұрын

In this scenario, there is not difference. The only situation requiring 'ln' is when the base of an index is 'e'.

@nothingbutpain863

Ай бұрын

Also, ln(x) is equivalent to log(e, x).

Even cooler is the fact that if you simplify the first answer they would be the same(I don't know if I'm right to be honest but iirc then that's cool) For example the numerator (ln 3 + 5 ln 2) can be rewritten as: ln 3 + ln 2⁵ or ln 3 +ln 32 Which is equal to ln (3•32) or ln 96 And the denominator can be rewritten as ln (2/3). Which means the equation can be written as [ ln 96/ln (2/3)] which is just equal to the second answer log base 2/3 of 96!

I'm an international relations major and somehow watched this whole video and nodded everytime he looked at me as if im getting everything he says

Could you not also do $$log_{2}3=(x+1)/(x-5)$$ ? (LaTeX format)

I figured immediately it was gonna be complicated because (although I could be wrong) there is no integer exponent of 2 that is divisible by 3

@Cibakro

3 ай бұрын

That's right -> Just express any exponent of 2 in terms of primes, you will never have the number 3. The same is true that no exponents of 3 are divisible by 2 for the same reason.

Watching this as a gcse student in the uk knowing this won’t come up in my exams but this was thoroughly interesting

What does the plate behind you contain? I like to collect things related to mathematics

i somehow went through algebra I and II, precalc, calc I and II, yet never saw any of this and now i feel like i was robbed. this looks so interesting and i am now lamenting never having had a math teacher that makes math interesting. thanks, random math guy on the internet!

@Roro-ej7ke

Ай бұрын

Same story🤷‍♀️

As a sophomore in College math, this actually makes sense

0:45 the quicker method after this step would be to divide x-5 by x+1, which would be equal to ln3/ln2. Now use componendo dividendo on both sides :)

What if we do 2^x=3 and then do log on both sides to get x log 2=log 3, log2/log3=x and then 2^x-5=2^(log2/log3)x+1 then do x-5=(log2/log3)x+1 and then work it out?

@expiredmilk5435

3 ай бұрын

X= log3/log2 not log2/log3

What would happen if we tried to first integrate or derivate both sides first, and then solve for X?

Nice liked that one

Thanks u verry much

You can solve this a lot quicker by just splitting up the exponents into 2^x, 2^-5, 3^x, and 3. Then isolating x is a matter of factoring it out of 3^x/2^x. Then you get log(1/(3*2^5)) with a logBASE of 3/2. The answer is -11.26

I always do the first way and it's simple for me.

But if you can use any log fir the first solution, how do I know that the result doesn't change depending on that

@somaannn

3 ай бұрын

because you take the same log on both sides of the equation. Like if you do +3 on both sides, it doesnt change anything since it cancels out.

I'm not sure why I'm clicking on this. I am an economics student and just reading about consumer behavior theory, and it contains Lagranian function, which I've never heard before and try to find wtf is that equation but anyway I'm satisfied with this video.

Am i still correct if i do the "prettier way" but i use ln 2/3 instead of log base 2/3??

@aheinrich2535

3 ай бұрын

Yes

Does (5.2log9) same?

Taking log on both sides we get (x-5)log2=(x+1)log3 now value of log 2 and log 3 with base e are 0.69 and 1.09 approx So 0.69x-3.45=1.09x+1.09 0.4x=-4.54 x=-4.54/0.4 which is -11.35 So the ans must be around -11.35!

The professor when ever I start copying the notes. 3:34

@stolenmonkey7477

3 ай бұрын

That made me laugh a bit lol

I just woke up in the afternoon,and a pg student of computer science,, still watching this before eating by breakfast

It's long. You can write it like (2^x)/32=3*3^x from where you can easily go to (2/3)^x=32*3

Idk why the heck im watching this in 3 year of high school but hell yeah this is so freaking cool

Not an approximation; the two answers are exactly the same. This we can see through a combination of the change of base formula and mainpulation of logs in the numerator and denominator of version #1: ln 3 + 5 ln 2 = ln 3 + ln 32 = ln 96; ln 2 - ln 3 = ln 2/3. As for a workable approximation, we should clean this up a bit first: for X = ln(96)/ln(2/3), we have X = -ln(96)/ln(3/2), and conveniently for log approximations, 3/2^2 ~ sqrt(5), 3/2^3 ~ sqrt(11). Now, 9870/9216 ~ 1.071, so we take 96^2 = 9216 ~ 9870 = 11^3 * sqrt(11) * sqrt(5) ~ (3/2)^23, 96 ~ (3/2)^23/2 for a final approximation of X = -11.5, which seems to agree OK with calculator results of ~-11.257.

@CuriousMindDevesh

2 ай бұрын

I exactly do this and get a error of approx. 0.25

I m a bio student why am watching this Great explanation 👍

Being a math tutor, at least around where I am, these days everybody's calculator is able to do logs with an arbitrary base, whether directly with the button or further inside a menu.

oh my fricking god how many whiteboard pen boxes do you have😂

When do we use ln and log in both sides?

@iancoffey2797

3 ай бұрын

@shrike1861 When the base of the exponents (here they are the 2 and 3) cannot be written as a power of either. e.g. if one side had 2^x and the other side had 32^x-1 we could rewrite the 32 as a power of 2, making it 2^5(x-1). You can then remove the common base and solve. That can't be done here so you must use log or ln

Learned logarithms 4 months ago, forgot them, then relearned them here

is 48 an option ?

Can't we simply do it like this? Applying log on both sides We have X-5)log2=x+1)log3 Now taking x+1 to the left side and log2 to right side. X-5/x+3=log3/log2 Applying division rule of log we get log3/log2=log (3-2) That is 0 So X-5/x+3=0 So X=5? Am a rookie at maths,so please tell me its correct or not

@ra1yms0up

2 ай бұрын

The log division rule is incorrect, and why x + 3 in the denominator?

@guywithcapacitors

2 ай бұрын

​@@ra1yms0up how is that incorrect, could you please tell And that x+3 is in the denominator cause it was multiplied in RHS I just got it to the LHS

@ra1yms0up

2 ай бұрын

@guywithcapacitors You can't turn log3/log2 to log(3-2), I think you meant log3-log2 = log(3/2) The incorrect rule you used works for exponents, so use them to rewrite an equation (I personally solved it like this): 2^(x-5) = (2^x)/(2^5) 3^(x+1) = (3^x)*3 We'll get: (2^x)/32 = (3^x)*3 Move 32 to the RHS and 3^x to the LHS: (2^x)/(3^x) = 32 * 3 32 * 3 = 96 x is an exponent for both 2 and 3, so you can take it out (2^x/3^x = (2/3)^x): (2/3)^x = 96 And that's it. You x is a log of 96, base 2/3 But I still can't understand where you got x+3 if there's x+1

@guywithcapacitors

2 ай бұрын

@@ra1yms0up ohhh thank you mate I really had messed up in that log part!😅 And yes that is x+1,so sorry from my part for putting that x+3😶‍🌫️

로그공식 까먹었던 걸 복습하게 되는 영상. 이거보고 다시 생각났네요

@snmnurr

25 күн бұрын

Hi in which grade do you learn logarithms in Korea?

@spspspd

25 күн бұрын

@snmnurr people learn logarithms in 11th grade in Korea, but Im now in high school in the US :)

Equivalent answer with slightly less distribution: 2^(x - 5) = 3^(x + 1) (x - 5) ln 2 = (x + 1) ln 3 x - 5 = (x + 1) log2(3) x - 5 = x log2(3) + log2(3) x - x log2(3) = 5 + log2(3) x = (5 + log2(3)) / (1 - log2(3))

2:25 imo you should have transposed the terms to other side respectively instead of subracting/adding on both the sides

@amitapatil1881

Ай бұрын

Anime Dekhne Vale life mae kuch nahi kar sakte

very nice.

Bro he the goat

Why not take log on both sides and apply Log(x)^n = nlogx And then reduce it to (x-5)log2=(x+1)log3

@unholyravioli989

3 ай бұрын

That's precisely what he did

@adithyakashyap5845

3 ай бұрын

Oh😂...

Why did you not use log table

Why not just get a log table and put values of log2 and log3

@violate6332

2 ай бұрын

It could come up on an exam where you cant use calculators or the log table

Took precalculus in high school roughly 20 years ago. NEVER learned this.

I played Mario kart tour every day during honors precalc in high school and I regret it every day as an engineering major in university

I'm a sophomore engineer major at a top university in the nation and had zero clue how to do this. I'm cooked

Can you solve this equation please? e^(x-1)/x =x I know that x equals 1, but I did not know how to prove it

What I like the best about this guy is he still uses a whiteboard and not a smart board

I don't know if anyone else has pointed this out, but you shouldn't have used x at the end when explaining how to use the change of base rule. It will definitely confuse people who are not very familiar with math as you have an equation for x right next to it on the board.

Damn, if you were my high school teacher I would have learned mathematical induction way before the week before my final exam

i just approximated to solve it quickly: x-5/x+1 = log3/log2 x-5/x+1 = 0.4771/0.3010 10x-50 = 16x+16 x=-11

My friends in my old algebra class had a funny way of remembering the ln(x^2)=2lnx theorem. We called it the yeet theorem because you take the exponent and yeet that shit to the front

You are great

It's crazy how i first saw the question in the thumbnail and knew that i had to add log on both sides

I got -11.256, but my calculations is a bit off bc i did it my way so wasnt sure how to get the exact number

Why not just take log base 2 of both sides in the beginning? It's more sensible than either option.

x = ln(96)/ln(2/3) + 2*pi*i/ln(2/3) * k where k is an integer. When k=0, you get a real answer.

My reaction if this was a question on the test: Answer test C Seems right

indians and asians are the only two people i trust to teach me maths. Indians for basics to intermediate and for advance i trust asians

that's way more complicated way that i was taught in school lmao

I'm not good at it, but I just love calculus and math now that I'm in college