Logarithm Problem

[mathmad2008]

Hi guys,

I'm having trouble with this problem:

3^(x + 4) = 6^(2x - 5)

Here's what I've done so far:

3^(x + 4) = 6^(2x - 5)
log( 6^(2x - 5) ) / log( 3 ) = x + 4
(2x - 5) * (log( 6 ) / log( 3 )) = x + 4
log( 6 ) / log( 3 ) = (x + 4) / (2x - 5)

Any help would be appreciated!

 

[Deleted member 4993]

Hi guys,

I'm having trouble with this problem:

3^(x + 4) = 6^(2x - 5)

Here's what I've done so far:

3^(x + 4) = 6^(2x - 5)
log( 6^(2x - 5) ) / log( 3 ) = x + 4
(2x - 5) * (log( 6 ) / log( 3 )) = x + 4
log( 6 ) / log( 3 ) = (x + 4) / (2x - 5)

Any help would be appreciated!

You are going in the correct direction.

How much is [log(6)/log(3)] ?

Then continue.....

 

[soroban]

Hello, mathmad2008!

$\displaystyle 3^{x + 4} \:=\: 6^{2x - 5}$

Take logs: .$\displaystyle \ln(3^{x+4}) \:=\:\ln(6^{2x-5})$

. . . . . . $\displaystyle (x+4)\ln3 \:=\: (2x-5)\ln6$

. . . . .$\displaystyle x\ln3 + 4\ln3 \:=\:2x\ln6 - 5\ln6$

. . . .$\displaystyle 2x\ln6 - x\ln3 \:=\: 4\ln3 + 5\ln6$

i . . .$\displaystyle (2\ln6 - \ln3)x \:=\:4\ln3 + 5\ln6$

. . . . . . . . . . . . . .$\displaystyle x \:=\:\dfrac{4\ln3 + 5\ln6}{2\ln6 - \ln3}$