Help with logs please

[lolilovepie]

Hi, I was wondering if youc an please help me solve this one problem. I started on it but then got stuck towards the end.

the problem is Solve. leave solution in exact form:

3^(x-2)=4^(2x+1)

==============

so this is what i did:

3^(x-2)=4^(2x+1)

=> log3(x-2)=log4(2x+1)

=>(x-2)=(log4/log3)(2x+1)

=> (x-2)=1.262 (2x+1)

=> (x-2)=2.24x + 1.262

and this is where i got stuck

 

[lolilovepie]

No, your approach isn't really valid past the 2nd line.

what you can do is use a common base for your logs on both sides.

\(\displaystyle \begin{align*}&\log(3^{x-2})==\log(4^{2x+1}) \\ \\

&(x-2)\log(3)==(2x+1)\log(4) \\ \\

&(\log(3)-2\log(4))x==\log(4)+2\log(3)\end{align*}\)

and I think you can solve it from here.

oh, how did you get (log(3)−2log(4))x==log(4)+2log(3)​ i'm kind of confuse at that part

 

[lolilovepie]

I just collected all the x terms together on the left and all the constant terms together on the right.

It's just a small bit of algebra.

ohh okay,
I did:
( x - 2 ) log 3 = ( 2x + 1 ) log 4 x log 3 - 2 log 3 = 2x log 4 + log 4
x log 3 - 2x log 4 = log 4 + 2 log 3
x ( log 3 - 2 log 4 ) = log 4 + 2 log 3
x = ( log 4 + 2 log 3 ) / ( log 3 - 2 log 4 )

and got x= -2.1?
not sure if that's right

 

[JeffM]

oh, how did you get (log(3)−2log(4))x==log(4)+2log(3)​ i'm kind of confuse at that part

oh, how did you get (log(3)−2log(4))x==log(4)+2log(3)​ i'm kind of confuse at that part

\(\displaystyle 3^{(x - 2)} = 4^{(2x + 1)} \implies\)

\(\displaystyle log \left(3^{(x - 2)}\right) = log \left(4^{(2x + 1)}\right) \implies\)

\(\displaystyle (x - 2)log(3) = (2x + 1)log(4) \implies\)

\(\displaystyle xlog(3) - 2log(3) = 2xlog(4) + log(4) \implies\)

\(\displaystyle xlog(3) - 2xlog(4) = 2log(3) + log(4) \implies\)

\(\displaystyle x\{log(3) - 2log(4)\} = 2log(3) + log(4)\implies\)

\(\displaystyle x = \dfrac{2log(3) + log(4)}{log(3) -2log(4)}.\) This is what romsek did. You can simplify further.

\(\displaystyle x = \dfrac{log(36)}{log(3/16)} \approx -2.14.\)