# Thread: logs problem giving trouble

1. ## logs problem giving trouble

$\displaystyle 3^{2-x}=2^{2x-1}$

2. Originally Posted by math321
$\displaystyle 3^{2-x}=2^{2x-1}$
$\displaystyle \log(3^{2-x}) = \log(2^{2x-1})$

$\displaystyle (2-x)\log{3} = (2x-1)\log{2}$

distribute both sides, get all terms with x on one side, then solve for x.

3. Hi Math321

This equation, when log laws are applied

$\displaystyle 3^{2-x}=2^{2x-1}$

gives you,

$\displaystyle (2-x)\ln(3)=(2x-1)\ln(2)$

now expand each side, group like terms and finish.

4. Originally Posted by math321
$\displaystyle 3^{2-x}=2^{2x-1}$
$\displaystyle 3^{2-x}=2^{2x-1}$

$\displaystyle (2-x)\ln{3}= (2x-1)\ln{2}$

$\displaystyle 2\ln{3}-x\ln{3}= 2x\ln{2}-\ln{2}$

$\displaystyle 2\ln{3}+\ln{2}= 2x\ln{2}+x\ln{3}$

$\displaystyle 2\ln{3}+\ln{2}= x(2\ln{2}+\ln{3})$

$\displaystyle \frac{2\ln{3}+\ln{2}}{2\ln{2}+\ln{3}}= x$

$\displaystyle x = \frac{2\ln{3}+\ln{2}}{2\ln{2}+\ln{3}} = \frac{\ln18}{\ln{12}}$

5. Hello, math321!

Solve for $\displaystyle x\!:\;\; 3^{2-x}\:=\:2^{2x-1}$

Take logs: .$\displaystyle \ln\left(3^{2-x}\right) \:=\:\ln\left(2^{2x-1}\right) \quad\Rightarrow\quad (2-x)\ln3 \:=\:(2x-1)\ln 2$

. . $\displaystyle 2\ln 3 - x\ln 3 \:=\:2x\ln2 - \ln 2 \quad\Rightarrow\quad 2x\ln 2 + x\ln 3 \:=\:2\ln3 + \ln 2$

. . $\displaystyle x(2\ln2 + \ln3) \:=\:2\ln3 + \ln 2 \quad\Rightarrow\quad x \:=\:\frac{2\ln 3 + \ln 2}{2\ln 2 + \ln 3}$

. . $\displaystyle x \:=\:\frac{\ln(3^2) + \ln 2}{\ln(2^2) + \ln3} \:=\:\frac{\ln(9\cdot2)}{\ln(4\cdot3)} \:=\:\frac{\ln18}{\ln12}$

Therefore: .$\displaystyle x \;\approx\;1.163171163$