Question

How do you solve \[5\left( {{2}^{3x}} \right)=8\]?

Answer 1

Hint: Divide both the sides with 8 and simplify the expression. Now, divide both the sides with 5 to remove the 5 from the L.H.S. Write 8 in the L.H.S. in exponential form having base equal to 2. Use the formula of ‘exponents and powers’ given as: - \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\] to simplify the L.H.S. further. Now, take log to the base 2, i.e., \[{{\log }_{2}}\], both the sides to form a linear equation in x. Solve for the value of x to get the answer.

Complete step by step answer:
Here, we have been provided with the exponential equation: - \[5\left( {{2}^{3x}} \right)=8\] and we are asked to solve it. That means we have to find the value of x.
\[\because 5\left( {{2}^{3x}} \right)=8\]
Dividing both the sides with 8, we get,
\[\Rightarrow \dfrac{5\left( {{2}^{3x}} \right)}{8}=1\]
Now, dividing both the sides with 5, we get,
\[\Rightarrow \left( \dfrac{{{2}^{3x}}}{{{2}^{3}}} \right)=\dfrac{1}{5}\]
Using the formula of ‘exponents and powers’ given as: - \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\], we get,
\[\begin{align}
  & \Rightarrow {{2}^{3x-3}}=\dfrac{1}{5} \\
 & \Rightarrow {{2}^{3\left( x-1 \right)}}=\dfrac{1}{5} \\
\end{align}\]
Taking log to the base 2, i.e., \[{{\log }_{2}}\], both the sides, we get,
\[\Rightarrow {{\log }_{2}}{{2}^{3\left( x-1 \right)}}={{\log }_{2}}\left( \dfrac{1}{5} \right)\]
Using the properties of logarithm: - \[\log {{a}^{m}}=m\log a\] and \[\log \left( \dfrac{1}{a} \right)=-\log a\] in the L.H.S. and R.H.S. respectively, we have,
\[\Rightarrow 3\left( x-1 \right){{\log }_{2}}2=-{{\log }_{2}}5\]
Using the formula: - \[{{\log }_{n}}n=1\] for n > 0 and \[n\ne 1\], we get,
\[\begin{align}
  & \Rightarrow 3\left( x-1 \right)\times 1=-{{\log }_{2}}5 \\
 & \Rightarrow 3\left( x-1 \right)=-{{\log }_{2}}5 \\
\end{align}\]
Dividing both the sides with 3, we have,
\[\begin{align}
  & \Rightarrow \left( x-1 \right)=\dfrac{-1}{3}{{\log }_{2}}5 \\
 & \Rightarrow x=1+\left( \dfrac{-{{\log }_{2}}5}{3} \right) \\
 & \Rightarrow x=1-\left( \dfrac{{{\log }_{2}}5}{3} \right) \\
\end{align}\]
Hence, the value of x is \[1-\left( \dfrac{{{\log }_{2}}5}{3} \right)\].

Note:
One may note that we cannot simplify the value of x further because we don’t know the value of \[{{\log }_{2}}5\]. However, we can use the base change rule given as \[{{\log }_{a}}b=\dfrac{{{\log }_{10}}b}{{{\log }_{10}}a}\] to simplify \[{{\log }_{2}}5\]. Now, either we need to use the log table to find the value of \[{{\log }_{10}}5\] and \[{{\log }_{10}}2\] or we need to memorize the log values of numbers from 1 to 10 for further simplification of \[{{\log }_{2}}5\].