Question

Solve the given equation for x:
${{3}^{x}}={{5}^{x-2}}$

Answer 1

- Hint: In the given equation, take log to the base 10 on both sides then we get $x\log 3=\left( x-2 \right)\log 5$ then solve this equation in x. The final expression of x in terms of logarithm we have $x=\dfrac{2}{1-{{\log }_{5}}3}$.

Complete step-by-step solution -

The equation given in the question is:
${{3}^{x}}={{5}^{x-2}}$
Taking log to the base 10 on both the sides we get,
$\log {{3}^{x}}=\log {{5}^{x-2}}$
There is a property of logarithm that the power in the log will come before the log.
$x\log 3=\left( x-2 \right)\log 5$
Rearranging the above equation by taking x on the one side of equal sign we get,
$2\log 5=x\left( \log 5-\log 3 \right)$
There is a property of logarithm that $\log a-\log b=\log \dfrac{a}{b}$ so using this relation in the above equation we get,
$2\log 5=x\log \dfrac{5}{3}$
We can also use the property of logarithm as $a\log b=\log {{b}^{a}}$ in the above equation.
$\log {{5}^{2}}=x\log \dfrac{5}{3}$
Rearranging x on the one side of the equation and log expressions on the other side we get,
$\dfrac{\log {{5}^{2}}}{\log \dfrac{5}{3}}=x$
Solving above logarithmic expression we get,
$\begin{aligned} & \dfrac{2}{\dfrac{\log 5-\log 3}{\log 5}} \\\ & =\dfrac{2}{1-\dfrac{\log 3}{\log 5}}=\dfrac{2}{1-{{\log }_{5}}3} \\\ \end{aligned}$
From the above calculations, we have solved the value of x in the given equation is:
$x=\dfrac{2}{1-{{\log }_{5}}3}$
Hence, the value of$x$in the given equation is $\dfrac{2}{1-{{\log }_{5}}3}$.

Note: You might think of getting the value of x by hit and trial method in the given equation.
${{3}^{x}}={{5}^{x-2}}$
For e.g. if we substitute $x=0$ in the above equation then we get,
$\begin{aligned} & 1={{5}^{-2}} \\\ & \Rightarrow 1=\dfrac{1}{25} \\\ \end{aligned}$
As you can see from the above that L.H.S ≠ R.H.S so $x=0$ is not the solution.
Now, if we substitute $x=1$ in the given equation we get,
$\begin{aligned} & {{3}^{1}}={{5}^{1-2}} \\\ & \Rightarrow 3={{5}^{-1}} \\\ \end{aligned}$
As you can see from the above that L.H.S ≠ R.H.S so $x=0$ is also not the solution.
Hence, the hit and trial method fails.
So, we should take log base 10 on both the sides in the given equation and then proceed.