How do you solve $3\left( {{5}^{x-1}} \right)=21$?

Hint: We first try to simplify the equation by dividing both sides of the equation $3\left( {{5}^{x-1}} \right)=21$ with 3. Then we try to take a logarithm with base as 5 to get the solution for $x$. We can also express the solution in only form of logarithm.

Complete step by step solution:
We have been given an indices problem where $3\left( {{5}^{x-1}} \right)=21$.
We first divide the equation on both sides with 3 and get
\[\begin{align}
  & \dfrac{3\left( {{5}^{x-1}} \right)}{3}=\dfrac{21}{3} \\
 & \Rightarrow \left( {{5}^{x-1}} \right)=7 \\
\end{align}\]
Now as the term 7 can’t be expressed as power form of 5, to solve the equation we need to take the help of logarithm.
We take logarithms with base 5.
Therefore, \[{{\log }_{5}}\left( {{5}^{x-1}} \right)={{\log }_{5}}7\].
Now we apply some logarithmic identities.
We have ${{\log }_{x}}{{a}^{p}}=p{{\log }_{x}}a$ and ${{\log }_{x}}x=1$.
Therefore, \[{{\log }_{5}}\left( {{5}^{x-1}} \right)=\left( x-1 \right){{\log }_{5}}5=\left( x-1 \right)\].
The final equation becomes \[\left( x-1 \right)={{\log }_{5}}7\].
We add 1 to the both sides of the equation \[\left( x-1 \right)={{\log }_{5}}7\] to find the solution of $x$.
\[\begin{align}
  & \left( x-1 \right)+1={{\log }_{5}}7+1 \\
 & \Rightarrow x={{\log }_{5}}7+1 \\
\end{align}\]
Therefore, the solution for $3\left( {{5}^{x-1}} \right)=21$ is \[{{\log }_{5}}7+1\].

Note: We can also express the solution of \[x={{\log }_{5}}7+1\] in a single form of logarithm. We know that ${{\log }_{x}}x=1$. We take $1={{\log }_{5}}5$. So, \[x={{\log }_{5}}7+{{\log }_{5}}5\].
We use the identity form of $\log a+\log b=\log \left( ab \right)$.
So, \[x={{\log }_{5}}7+{{\log }_{5}}5={{\log }_{5}}\left( 5\times 7 \right)={{\log }_{5}}35\].
Therefore, the solution for $3\left( {{5}^{x-1}} \right)=21$ can be expressed as \[{{\log }_{5}}35\].