The backing up speed in the beginning though lol

@blast8012

2 жыл бұрын

Sorry im slow... how did he get 3 to the power of log 6 base 3...

@wz_lx24

2 жыл бұрын

@@blast8012 Whenever you have a log with a base, its value is the solution for x: a^x = b; log_a(b) = x So if you take log_3(6) it's the solution for x in this equation: 3^x = 6; x = log_3(6) So if you wanna get 6, and 6 is equal to 3^x, and x is equal to log_3(6), 6 is equal to 3^(log_3(6)).

Thanks, I'm re-learning algebra. You're lessons are helpful! I needed it

It is very simple 3^(x)½=6^(x²) We take the logarithm of both sides. (X½)Log(3)=x²(log(6) Move to one side. Factoring from x½ X½[log(3)-x^(3/2)log(6)]=0 So x½=0 x=0 And x=(log(3)/(log(6))⅔

@alrycaaeveahexendias1236

2 жыл бұрын

You lost me at "take the logarithm of both sides". Sorry for the skepticism, as it looked like a leap of faith from my perspective. How did that maneuver NOT break the equation?

@morteza3268

2 жыл бұрын

@@alrycaaeveahexendias1236 If you choose the right starting point, the path to the goal will be shortened.

@marcmarco1305

Жыл бұрын

Nice

@kockorzo

10 ай бұрын

Ah yes, since 2^6=4^3 we can take the log of both sides giving us log2(64)=log4(64) I get that you’re joking but it’s a poor joke since it might legitimately reinforce some people with false understanding.

@duccline

7 ай бұрын

​@@alrycaaeveahexendias1236 doing the same thing to both sides of the equation maintains the equality, because (as expressed by the = sign) the values on both sides are necessarily equal

that was great! Really brought up some important issues when solving

I love this trick! NICE!

By taking log on both sides, it can be solved

Nice one!

As a teacher do you give these type of questions to your students lol ? In my school (EU), We would never ever face such problems that requires you to actually think deeper... Anyway love the channel !

@stewbaka4279

2 жыл бұрын

bro just left you on read 😭😭😭😭

@junkgum

Жыл бұрын

#YTAlgebra

@marcmarco1305

Жыл бұрын

I’m from EU too..we have never been given these type of equations😂😂 but these are really easy once you understand the trick tbh

@Iamblindanddeaf23

Жыл бұрын

@@marcmarco1305 I mean yeah in general they train us to compute stuff over and over again, especially with logarithms, you legit feels like a bot

Here's my solution: 3^sqrt(x) = 6^(x²) sqrt(x)ln(3) = x²ln(6) x^(3/2) = ln(3)/ln(6) (assuming x =/= 0) x = [log_6(3)]^(2/3) I think this is more legitimate than factoring a zero, since that method assumes the function is continous.

@yesveryprofesionalnameyes6055

Жыл бұрын

dividing by zero generally is a worse method than factorisation because you remove a solution, so you also gotta plug in for x = 0 (when it's that obvious, but in some more complex cases it might not be). Factoring just assures you that it's a solution. Showing continuity isn't that hard here because log_n(x): R+ --> R (for the real-valued logarithm where n is positive) and everything else is a polynomial

Question when you changed the base of 6 to 3^(log6/log3) what is that concept called and where can I learn about it?

@pedrosso0

2 жыл бұрын

simply change of base. I don't know where you can learn about it but it is fairly simple. a = e^(ln a) = e^(ln a ln b / ln b) = e^(ln b))^(ln a / ln b) = b^(ln a / ln b) therefore a = b^(ln a / ln b) should work forall non zero a and b, so long as b isn't 1 (that is, for real values)

@rockybruno9316

2 жыл бұрын

Property of Equality for Exponential Equations- this property is useful to solve an exponential equation with the same bases. It says when the bases on both sides of an exponential equation are equal, then the exponents must also be equal. You learn this concept on khan Academy or other youtube videos.

@REV0LUT10N_

Жыл бұрын

the concept is that log a to the base b = log a/log b and log a to the base b=1/log b to the base a

You could get the log of both sides from the beginning and it would have been simpler.

(log(6)(3))2/3

Where is that other video?

awesome

1:48 At this point, instead of squaring both sides, can we multiply both sides by sqrt(x)?

@zahranf.a.9864

2 жыл бұрын

Then how to do next step?

@aaditrangnekar

Жыл бұрын

that may complicate it cause it gonna be sone sqrt(x) . (x^2 .....)

@alejrandom6592

Жыл бұрын

Yes you can, but it doesn't help much

Best intro

Love frm India 🇮🇳 ❤

Le soluzioni sono x=0,x=(log3)^2/3,con log in base 6

amazing!!! still too easy haha

Nice videoo

1dlate

No fair making us wait for the other video

i lv u 99.9%

2:29 but after dividing by x the solution x=0 is ruled out

@saar4869

2 жыл бұрын

we know that x = 0 is a solution so it doesn't matter that we won't get it in the end

@alejrandom6592

Жыл бұрын

He said that

You don't have to check the solutions since both sides of the equation you squared were non-negative.

Shouldn't there be 4 solutions? Since we had a quartic equation and then factored it

At first second I thought I'm watching a spin-off to Tenet.

I tried to solve it, and I came up with: x = [(log2 3) / (1 + log2 3)]^(1/3). Does it work too?

@ProfAmeen08

2 жыл бұрын

AHHHH! SHIT! I must've been mistaken; it doesn't work.

@marcmarco1305

Жыл бұрын

@@ProfAmeen08 at least you tried..have you figure it out?

@ProfAmeen08

Жыл бұрын

@@marcmarco1305 Yeah, I guess I knew where my mistake was; thx.

@MichaelRothwell1

Жыл бұрын

The power 1/3 should be 2/3, otherwise it's OK. Note that 1+log₂3=log₂2+log₂3=log₂6, so your fraction = log₂3/log₂6=log3/log6, same as answer in video.

now demonstrate any complex solution!

By a look, we can say that 0 is definitely a solution.

x^3 = something. 3 roots. So, where is 2 complex friends

why does he sound so unusually unenergetic????

* multiplies both sides by 0* *leaves the class* This a joke

0

isnt x just 0

Why so complicated? By inspection, x = 0 is an obvious root. Now consider only nonzero roots. Take logs (to any consistent base) on both sides: sqrt x log 3 = x^2 log 6. Since x not equal to zero by assumption, rearrange to x^(3/2) = log3/log6 So x = (log3/log6)^(2/3), which can also be written more compactly (but less usefully) as [log_6 (3)] ^(2/3) - the subscript means log to base 6. So the two solutions are x = 0 and x = [log_6 (3)] ^(2/3) = (log3/log6)^(2/3)

DON T