Question

How do you solve \[{{25}^{2x+1}}=144\]?

Answer 1

Hint: In the given equation \[{{25}^{2x+1}}=144\], the variable term is present as an exponent on the left hand side. So it cannot be solved by using the basic algebraic operations. For solving, we need to remove the variable from the exponent. For this we have to take the logarithm of both the sides of the equation. Then using the property $\log {{a}^{m}}=m\log a$, we will be able to remove the variable term from the exponent. Then we will obtain a linear equation in $x$ which can be solved by using the basic algebraic operations.

Complete step by step solution:
The equation given in the above question is
\[\Rightarrow {{25}^{2x+1}}=144\]
Since the variable is present as an exponent, we cannot solve it directly. So in order to simplify the given equation, we take the logarithm on both the sides to get
\[\Rightarrow \log \left( {{25}^{2x+1}} \right)=\log 144\]
Now, from the properties of the logarithm function we know that $\log {{a}^{m}}=m\log a$. Applying this property on the LHS of the above equation, we get
$\Rightarrow \left( 2x+1 \right)\log 25=\log 144$
Dividing both the sides by $\log 25$ we get
$\Rightarrow 2x+1=\dfrac{\log 144}{\log 25}$
Subtracting $1$ from both the sides we get
\[\begin{align}
  & \Rightarrow 2x=\dfrac{\log 144}{\log 25}-1 \\
 & \Rightarrow 2x=\dfrac{\log 144-\log 25}{\log 25} \\
\end{align}\]
From the properties of logarithm we also know that $\log A-\log B=\log \dfrac{A}{B}$. So the RHS of the above equation becomes
$\begin{align}
  & \Rightarrow 2x=\dfrac{\log \left( \dfrac{144}{25} \right)}{\log 25} \\
 & \Rightarrow 2x=\dfrac{1}{\log 25}\log \left( \dfrac{144}{25} \right) \\
\end{align}$
Finally, dividing both the sides by $2$ we get
$\Rightarrow x=\dfrac{1}{2\log 25}\log \left( \dfrac{144}{25} \right)$
Hence, the given equation is solved.

Note: We may be interested in further simplifying the final solution obtained by writing $144={{12}^{2}}$, $25={{5}^{2}}$ and applying the logarithm property $\log {{a}^{m}}=m\log a$ again to obtain the final solution as \[x=\dfrac{1}{2\log 5}\log \left( \dfrac{12}{5} \right)\]. But this is not a simplified form. It is only a reduced expression. So we have not used this in the above solution.