Question

If \[{{10}^{2y}}=25\], then \[{{10}^{-y}}\] equals to,
A. $-\dfrac{1}{5}$
B. $\dfrac{1}{50}$
C. $\dfrac{1}{625}$
D. $\dfrac{1}{5}$

Answer 1

Hint: We will start by taking log base 10 on both side of the equation \[{{10}^{2y}}=25\] and using the logarithm identity that ${{a}^{\log b}}={{b}^{\log a}}$ to find the value of \[{{10}^{-y}}\] for y. We will use the logarithm identity that $\log {{a}^{b}}=b\log a$.

Complete step-by-step answer:
Now, we have been given that \[{{10}^{2y}}=25\].
Now, taking log base 10 on both sides we have,
\[\begin{align}
  & \log {{10}^{2y}}=\log 25 \\
 & 2y{{\log }_{10}}10={{\log }_{10}}{{5}^{2}} \\
\end{align}\]
Now, we know that,
$\begin{align}
  & {{\log }_{10}}10=1 \\
 & \log {{a}^{b}}=b\log a \\
\end{align}$
So, we have,
$\begin{align}
  & 2y=2{{\log }_{10}}5 \\
 & y={{\log }_{10}}5..........\left( 1 \right) \\
\end{align}$
Now, we have to find the value of \[{{10}^{-y}}\]. So, from (1) we have,
\[{{10}^{-y}}=10-{{\log }_{10}}^{5}\]
Now, we know that,
$\begin{align}
  & -{{\log }_{10}}a={{\log }_{10}}{{a}^{-1}} \\
 & -{{\log }_{10}}a={{\log }_{10}}\left( \dfrac{1}{a} \right) \\
 & {{b}^{\log a}}={{a}^{\log b}} \\
 & \Rightarrow 10-{{\log }_{10}}5=10{{\log }_{10}}{{5}^{-1}} \\
 & =10{{\log }_{10}}\dfrac{1}{5} \\
 & =\left( \dfrac{1}{5} \right){{\log }_{10}}10 \\
 & =\dfrac{1}{5} \\
\end{align}$
Hence, the correct option is (D).

Note: It is important to note that we have used fact that
\[\begin{align}
  & {{a}^{\log b}}={{b}^{\log a}} \\
 & -{{\log }_{10}}a={{\log }_{10}}\dfrac{1}{a} \\
\end{align}\]
Also, it is important to note that to find the value of \[{{10}^{-y}}\]. We have used the identity that ${{a}^{\log b}}={{b}^{\log a}}$. This identity is valid for all valid bases of log not only 10. As there are a lot of identities being used, we must be alert and careful not to make any silly mistakes.