Question

How do you solve ${{25}^{x}}={{125}^{x-2}}$ ?

Answer 1

Hint: We first convert all the numerical terms to $5$ in order to attain similarity on both the sides. We know that $25=5\times 5$ and $125=5\times 5\times 5$ . The LHS becomes ${{5}^{2x}}$ and the RHS becomes ${{5}^{3\left( x-2 \right)}}$ . The indices can be equated as $2x=3x-6$ , which can be solved to get $6$ .

Complete step by step answer:
The given equation that we have at this moment is,
${{25}^{x}}={{125}^{x-2}}$
This problem requires the concept of powers or indices. In order to solve this problem of indices or powers, we have to make the base of the powers equal or comparable. If we observe the above equation closely, we can see that $125$ is greater than $25$ and it is a multiple of $25$ as well. So, the bases can be made equal or comparable if we express $125$ as $25$ times something. We know that $125=25\times 5$ . Incorporating this in the above equation, we get,
$\Rightarrow {{25}^{x}}={{\left( 25\times 5 \right)}^{x-2}}$
But, there still remains an extra term $5$ . We need to get rid of it. But, instead of getting rid of it, if we convert all the terms to $5$ , it will be better. We know that, $25=5\times 5$ . So, the equation becomes,
$\Rightarrow {{\left( 5\times 5 \right)}^{x}}={{\left( 5\times 5\times 5 \right)}^{x-2}}$
The above equation can be simplified as,
$\Rightarrow {{\left( {{5}^{2}} \right)}^{x}}={{\left( {{5}^{3}} \right)}^{x-2}}$
We know that, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . So,
$\Rightarrow {{5}^{2x}}={{5}^{3\left( x-2 \right)}}$
This becomes,
$\Rightarrow {{5}^{2x}}={{5}^{3x-6}}$
We know that, if the bases on two sides are the same, we can equate their exponents. So,
$\Rightarrow 2x=3x-6$
We bring the $3x$ term to the LHS and get,
$\begin{align}
  & \Rightarrow 2x-3x=-6 \\
 & \Rightarrow -x=-6 \\
 & \Rightarrow x=6 \\
\end{align}$

Therefore, we can conclude that the answer of the problem is $6$.

Note: Students may at first seem this problem to be unsolvable. But this can be done easily if we take the above approach. This problem can also be solved by taking logarithms. If we take logarithms, we get,
$\begin{align}
  & x\log 25=\left( x-2 \right)\log 125 \\
 & \Rightarrow 2x\log 5=3\left( x-2 \right)\log 5 \\
 & \Rightarrow 2x=3x-6 \\
 & \Rightarrow x=6 \\
\end{align}$