Solving Exponential Equation different base(non common)

**jjtjp** · February 21st 2013, 06:29 PM

Here's the problem:
$2^{2x+12}= 3^{x-34}$

I can get it to this by taking the log of both sides:
$\frac{2x+12}{x-34}=\frac{log(3)}{log(2)}$

Now I could just use a calculator and get the value for the right side and the clean it up. But how would I finish this problem and get the answer in terms of log or ln. Can someone give me a hand? Thanks!

**ibdutt** · February 21st 2013, 07:31 PM

Just cross multiply and collect terms containing x on one side. You will have
2x log 2 + 12 log 2 = x log 3 - 34 log 3
x log 4 + 12 log 2 = x log 3 - 34 log 3
x log 4 - x log 3 = - 12 log 2- 34 log 3
x log 4/3 = - 12 log 2- 34 log 3 etc.

**Soroban** · February 21st 2013, 09:07 PM

Hello, jjtjp!

$\text{Solve for }x\!:\;\;2^{2x+12}\:=\: 3^{x-34}$

I can get it to this by taking the log of both sides:

$\frac{2x+12}{x-34}=\frac{\log(3)}{\log(2)}$
This is correct, but why did you create fractions?

We have: . $2^{2x+12} \:=\:3^{x-34}$

Take logs: . $\ln\left(2^{2x+12}\right) \:=\:\ln\left(3^{x-34}\right)$

. . . . . . . $(2x+12)\ln2 \:=\:(x-34)\ln3$

n . . . . $2x\ln2 + 12\ln2 \:=\:x\ln3 - 34\ln3$

. . . . . . $2x\ln2 - x\ln3 \:=\:-34\ln3 - 12\ln 2$

. . . . . . $x(2\ln2 - \ln3) \:=\:-(34\ln3 + 12\ln2)$

. . . . . . . . . . . . . . . $x \:=\:-\frac{34\ln3 + 12\ln2}{2\ln2-\ln3}$

. . . . . . . . . . . . . . . $x \:\approx\: -158.7536672$

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