Question

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

Answer 1

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

& \dfrac{3\left( {{5}^{x-1}} \right)}{3}=\dfrac{21}{3} \\\ & \Rightarrow \left( {{5}^{x-1}} \right)=7 \\\ \end{aligned}$$ Now as the term 7 can’t be expressed as power form of 5, to solve the equation we need to take the help of logarithm. We take logarithms with base 5. Therefore, $${{\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{aligned} & \left( x-1 \right)+1={{\log }_{5}}7+1 \\\ & \Rightarrow x={{\log }_{5}}7+1 \\\ \end{aligned}$$ **Therefore, the solution for $3\left( {{5}^{x-1}} \right)=21$ is $${{\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$$.