Question

How do you solve \[{15^{2x}} = 36\] ?

Answer 1

Hint:We can clearly see that the base on both the sides is not the same. Therefore, we will take logarithms on both the sides. After that, we will use the power rule of the logarithm. And finally, we will solve the equation for $x$.

Complete step by step answer:
We are given \[{15^{2x}} = 36\]. Here, we will apply the following procedure to solve this equation: Keep the exponential expression by itself on one side of the equation.Get the logarithms of both sides of the equation. You can use any bases for logs. Solve for the variable and keep the answer exact or give decimal approximations. As $2x$ is the power on the left hand side and the base on both sides is different. Therefore, we will now take logarithm on both the sides.
\[{15^{2x}} = 36 \\
\Rightarrow {\log _{10}}{15^{2x}} = {\log _{10}}36 \\ \]
Now, we will use the power rule or exponential rule for logarithm which states that the logarithm of a number with a rational exponent is equal to the exponent times its logarithm.
Therefore, we can write
\[{\log _{10}}{5^{2x}} = 2x{\log _{10}}15\]
Thus, we have
\[{15^{2x}} = 36 \\
\Rightarrow {\log _{10}}{15^{2x}} = {\log _{10}}36 \\
\Rightarrow 2x{\log _{10}}15 = {\log _{10}}36 \\ \]
Now, we will solve the equation for $x$.
\[{15^{2x}} = 36 \\
\Rightarrow {\log _{10}}{15^{2x}} = {\log _{10}}36 \\
\Rightarrow 2x{\log _{10}}15 = {\log _{10}}36 \\
\therefore x = \dfrac{{{{\log }_{10}}36}}{{2{{\log }_{10}}15}} \approx 0.66 \]
Thus, our final answer is $0.66$.

Note:Here, we have solved the equation by taking the log with base 10. However, we can also solve it by taking log with base 15.
\[{15^{2x}} = 36 \\
\Rightarrow {\log _{15}}{15^{2x}} = {\log _{15}}36 \\
\Rightarrow 2x{\log _{15}}15 = {\log _{15}}36 \\ \]
We know that when the number and the base of logarithm is the same, the result will be 1.
Therefore, we can take ${\log _{15}}15 = 1$.
We will now use this value in the main equation.
\[{15^{2x}} = 36 \\
\Rightarrow {\log _{15}}{15^{2x}} = {\log _{15}}36 \\
\Rightarrow 2x{\log _{15}}15 = {\log _{15}}36 \\
\Rightarrow 2x \times 1 = {\log _{15}}36 \\
\therefore x = \dfrac{{{{\log }_{15}}36}}{2} \approx 0.66 \]
Thus, by using both the methods, we get the same answer.