Question

How do you solve: \[2{{\left( 7 \right)}^{4x-5}}=30\]?

Answer 1

Hint: Divide both the sides of the given exponential equation with 2 to simplify. Now, take log to the base 7 on both sides and use the formula: - \[\log {{a}^{m}}=m\log a\] to simplify the L.H.S. In the R.H.S. write the obtained number as the product of its primes and use the formula: - \[\log \left( mn \right)=\log m+\log n\]. Solve the linear equation in x to get the answer.

Complete step by step solution:
Here, we have been provided with the exponential equation: - \[2{{\left( 7 \right)}^{4x-5}}=30\] and we are asked to solve it. That means we have to find the values of x.
\[\because 2{{\left( 7 \right)}^{4x-5}}=30\]
Now, dividing both the sides with 2 and simplifying by cancelling the common factors, we get,
\[\Rightarrow {{\left( 7 \right)}^{4x-5}}=15\]
Here, we need to remove the vase 7 in the L.H.S. To do this, let us take log to the base 7 on both the sides, so we have,
\[\Rightarrow {{\log }_{7}}\left[ {{7}^{4x-5}} \right]={{\log }_{7}}15\]
Using the identity \[\log {{a}^{m}}=m\log a\] to simplify the L.H.S., we get,
\[\Rightarrow \left( 4x-5 \right){{\log }_{7}}7={{\log }_{7}}15\]
Now, we know that when the base and argument of log is the same then its value is 1. The value of base must not be 1 and both the base and argument must be greater than 0. So, we have,
\[\begin{align}
  & \Rightarrow \left( 4x-5 \right)\times 1={{\log }_{7}}15 \\
 & \Rightarrow 4x-5={{\log }_{7}}15 \\
\end{align}\]
Here, we can write \[15=3\times 5\] as the product of its primes, so we have,
\[\Rightarrow 4x=5+{{\log }_{7}}\left( 3\times 5 \right)\]
Using the formula: - \[\log \left( mn \right)=\log m+\log n\], we get,
\[\Rightarrow 4x=5+{{\log }_{7}}3+{{\log }_{7}}5\]
Dividing both the sides with 4, we get,
\[\Rightarrow x=\dfrac{1}{4}\left[ 5+{{\log }_{7}}3+{{\log }_{7}}5 \right]\]
Now, using the base change formula of logarithm given as: - \[{{\log }_{a}}b=\dfrac{{{\log }_{c}}b}{{{\log }_{c}}a}\], we can change the above logarithmic expression having base 7 into logarithmic expression having base 10. So, we have,
\[\begin{align}
  & \Rightarrow x=\dfrac{1}{4}\left[ 5+\dfrac{{{\log }_{7}}3}{\log 7}+\dfrac{{{\log }_{7}}5}{\log 7} \right] \\
 & \Rightarrow x=\dfrac{1}{4}\left[ 5+\left( \dfrac{\log 3+\log 5}{\log 7} \right) \right] \\
\end{align}\]

Note: One may note that we cannot simplify the value of x further without knowing the values of log 3, log 5 and log 7. To find their values we have to memorize them or we need to use the log table. We have changed the base of the log as 10 because generally we memorize or find the values of log to the base 10 of a given number using log table. You may remember: - log 3 = 0.48, log 5 = 0.7 and log 7 = 0.84.