Question

How do you solve $3{{\left( 5 \right)}^{2x-3}}=6$?

Answer 1

Hint: First of all, we need to separate the variable and the constant terms in the given equation $3{{\left( 5 \right)}^{2x-3}}=6$. For this, we have to divide both the sides of the equation by $3$ after which the equation will get reduced to ${{\left( 5 \right)}^{2x-3}}=2$. Then, to take out the variable exponent, we need to take the logarithm on both the sides of the equation to get $\left( 2x-3 \right)\log 5=\log 2$ using the logarithm property which is given by $\log {{a}^{m}}=m\log a$. The obtained equation can then be easily solved using the basic algebraic manipulations and the properties of the logarithm function to get the required solution of the given equation.

Complete step by step solution:
The equation given to us in the above question is written as
$\Rightarrow 3{{\left( 5 \right)}^{2x-3}}=6$
Firstly, in order to separate the variable and the constant terms, we divide both the sides of the above equation by $3$ to get
\[\begin{align}
  & \Rightarrow \dfrac{3{{\left( 5 \right)}^{2x-3}}}{3}=\dfrac{6}{3} \\
 & \Rightarrow {{\left( 5 \right)}^{2x-3}}=2 \\
\end{align}\]
Now, taking the logarithm on both the sides of the above equation, we get
$\Rightarrow \log {{\left( 5 \right)}^{2x-3}}=\log \left( 2 \right)$
Using the property of the logarithmic function which is given by $\log {{a}^{m}}=m\log a$, we can write the above equation as
$\Rightarrow \left( 2x-3 \right)\log \left( 5 \right)=\log \left( 2 \right)$
Now, on dividing both the sides of the above equation by $\log \left( 5 \right)$, we get
\[\begin{align}
  & \Rightarrow \dfrac{\left( 2x-3 \right)\log \left( 5 \right)}{\log \left( 5 \right)}=\dfrac{\log \left( 2 \right)}{\log \left( 5 \right)} \\
 & \Rightarrow 2x-3=\dfrac{\log \left( 2 \right)}{\log \left( 5 \right)} \\
\end{align}\]
Adding \[3\] on both the sides, we get
$\begin{align}
  & \Rightarrow 2x-3+3=\dfrac{\log \left( 2 \right)}{\log \left( 5 \right)}+3 \\
 & \Rightarrow 2x=\dfrac{\log \left( 2 \right)}{\log \left( 5 \right)}+3 \\
\end{align}$
Taking LCM on the RHS, we get
$\Rightarrow 2x=\dfrac{\log \left( 2 \right)+3\log \left( 5 \right)}{\log \left( 5 \right)}$
Using the logarithm property $m\log a=\log {{a}^{m}}$ we can write
\[\begin{align}
  & \Rightarrow 2x=\dfrac{\log \left( 2 \right)+\log {{\left( 5 \right)}^{3}}}{\log \left( 5 \right)} \\
 & \Rightarrow 2x=\dfrac{\log \left( 2 \right)+\log \left( 125 \right)}{\log \left( 5 \right)} \\
\end{align}\]
Now, using the logarithmic property given by $\log A+\log B=\log \left( AB \right)$ we can write
$\begin{align}
  & \Rightarrow 2x=\dfrac{\log \left( 2\times 125 \right)}{\log \left( 5 \right)} \\
 & \Rightarrow 2x=\dfrac{\log \left( 250 \right)}{\log \left( 5 \right)} \\
\end{align}$
Finally, dividing the above equation by $2$ we get
\[\begin{align}
  & \Rightarrow \dfrac{2x}{2}=\dfrac{\log \left( 250 \right)}{2\log \left( 5 \right)} \\
 & \Rightarrow x=\dfrac{\log \left( 250 \right)}{2\log \left( 5 \right)} \\
\end{align}\]
Again using the logarithmic property $m\log a=\log {{a}^{m}}$, we can simplify the denominator as
$\begin{align}
  & \Rightarrow x=\dfrac{\log \left( 250 \right)}{\log {{\left( 5 \right)}^{2}}} \\
 & \Rightarrow x=\dfrac{\log \left( 250 \right)}{\log \left( 25 \right)} \\
\end{align}$

Hence, we obtained the solution for the given equation as $\dfrac{\log \left( 250 \right)}{\log \left( 25 \right)}$.

Note: We must remember all the important properties of the logarithmic function for solving these types of equations, where the variable is an exponent. We can also apply the exponent property ${{a}^{m-n}}=\dfrac{{{a}^{m}}}{{{a}^{n}}}$ to simplify the equation as $3\left( \dfrac{{{5}^{2x}}}{{{5}^{3}}} \right)=6$. This way, we will be able to get the solution quickly.