# How do you solve 3^(2-x)=5^(2x+1)?

Dec 24, 2015

Exponential rules and use of power rule of logarithm can help solve such a problem easily. The step by step working with explanation is given below.

#### Explanation:

${3}^{2 - x} = {5}^{2 x + 1}$
Splitting

${3}^{2} \cdot {3}^{-} x = {5}^{2 x} \cdot {5}^{1}$ Note: ${a}^{m + n} = {a}^{m} \cdot {a}^{n}$
$9 \cdot {3}^{-} x = {\left({5}^{2}\right)}^{x} \cdot 5$ Note : $\left({a}^{m n}\right) = {\left({a}^{m}\right)}^{n}$

$\frac{9}{3} ^ x = {25}^{x} \cdot 5$ Note : ${a}^{- m} = \frac{1}{a} ^ m$

Solving for x by isolating the term containing x to one side of the equation.

Multiply by ${3}^{x}$ on both sides we get.

$9 = {25}^{x} \cdot 5 \cdot {3}^{x}$

Dividing by 5 on both sides we get.

$\frac{9}{5} = {25}^{x} \cdot {3}^{x}$
$\frac{9}{5} = {\left(25 \cdot 3\right)}^{x}$
$\frac{9}{5} = {\left(75\right)}^{x}$

Taking log on both the sides

$\log \left(\frac{9}{5}\right) = \log {\left(75\right)}^{x}$
$\log \left(\frac{9}{5}\right) = x \log \left(75\right)$

Dividing $\log \left(75\right)$ on both the sides.

$\log \frac{\frac{9}{5}}{\log} \left(75\right) = x$

$x = \log \frac{\frac{9}{5}}{\log} \left(75\right)$ answer

Note: The answer can be represented in many ways and that would be decided by the question. If a numerical value is needed then please use a calculator to find it.

$x = {\log}_{75} \left(\frac{9}{5}\right)$ can be one answer as well. Make a choice depending on the question.