Question

How do you solve ${{2}^{\log x}}=\dfrac{1}{4}$?

Answer 1

Hint: First simplify $\dfrac{1}{4}$ in terms of power of ‘2’. Then ${{\log }_{2}}$ on both the sides. Use the formulae like $\log {{m}^{n}}=n\log m$ and ${{\log }_{a}}a=1$ for further simplification. Then raise both the sides by the power of ‘e’ to eliminate the logarithm part. Do necessary calculations to find the value of ‘x’.

Complete step by step solution:
To solve ${{2}^{\log x}}=\dfrac{1}{4}$, we have to find the value of ‘x’ for which the equation gets satisfied.
The equation we have ${{2}^{\log x}}=\dfrac{1}{4}$
The right side of the equation can be simplified as $\dfrac{1}{4}=\dfrac{1}{{{2}^{2}}}={{2}^{-2}}$
Now the equation becomes
$\Rightarrow {{2}^{\log x}}={{2}^{-2}}$
Taking ${{\log }_{2}}$ both the sides, we get
$\Rightarrow {{\log }_{2}}\left( {{2}^{\log x}} \right)={{\log }_{2}}\left( {{2}^{-2}} \right)$
As we know $\log {{m}^{n}}$ can be written as $\log {{m}^{n}}=n\log m$
So, ${{\log }_{2}}\left( {{2}^{\log x}} \right)=\left( \log x \right)\left( {{\log }_{2}}2 \right)$ and ${{\log }_{2}}\left( {{2}^{-2}} \right)=-2\left( {{\log }_{2}}2 \right)$
Hence, our equation becomes
$\Rightarrow \left( \log x \right)\left( {{\log }_{2}}2 \right)=-2\left( {{\log }_{2}}2 \right)$
Again as we know ${{\log }_{a}}a=1$, so ${{\log }_{2}}2=1$
Putting the value of ${{\log }_{2}}2$, we get
$\begin{align}
  & \Rightarrow \log x\times 1=-2\times 1 \\
 & \Rightarrow \log x=-2 \\
\end{align}$
Raising both the side by base power ‘e’, we get
$\Rightarrow {{e}^{\log x}}={{e}^{-2}}$
Cancelling out ‘e’ to the power ‘log’ on the left side of the equation and writing ${{e}^{-2}}$ as $\dfrac{1}{{{e}^{2}}}$ on the right side of the equation, we get
$\Rightarrow x=\dfrac{1}{{{e}^{2}}}$
This is the required solution of the given question.

Note: First $\dfrac{1}{4}$ should be written in terms of power of ‘2’ to convert the given equation to a form where various logarithmic formulae can be applicable for further simplification. During the simplification we have taken the base of log as ‘e’, generally It can also be taken as 10 as there is nothing mentioned in the question.