# Thread: A rather basic log question I believe

1. ## A rather basic log question I believe

"Solve for x:"

2 X 3^2x+3 = 3 X 2^3x+4

(X is multiply)

feel silly for posting this, is a simple log question i think, but its got me stumped.

thanks

2. Just to clarify, you want to solve $\displaystyle 2\cdot 3^{2x}+3=3\cdot 2^{3x}+4.$ Is that correct? If so, what steps have you taken?

3. it could be also:
$\displaystyle 2(3)^{2x+3} = 3(2)^{3x+4}$

in which case $\displaystyle x = -1$

4. sorry yeah the above is the way it's laid out - how did you get -1 ?

5. ## well several ways to solve this..

well several ways to solve this..

but you can divide both sides by $\displaystyle 2^1$ and $\displaystyle 3^1$ to elimate the coeficients,

then
$\displaystyle 3^{2x+2} = 2^{3x+3} \Rightarrow (2x+2)\ln{3} = (3x+3)\ln{2} \Rightarrow \frac{2x+2}{3x+3} = \frac{\ln{2}}{\ln{3}}$
after calculation we are left with

$\displaystyle x(.107) = -.107\Rightarrow x= -1$

6. ahhh it's so simple, i'm so stupid. haha thanks, i just didn't start with the basic, "dividng by 2" part. oops.

7. 2(3)^2x+3 = 3(2)^3x+4

log 2 × log 3^(2x+3) = log 3 × log 2^(3x+4)

(2x+3) log 3 = log 3
(3x+4) log 2 = log 2

2x log 3 + 3 log = log 3

2x log 3 = log 3 - 3 log 3

x(2 log 3) = -2 log 3

x = -2 log 3
/2 log 3

x = -1

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