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How do you solve ${0.5^x} = {16^2}$?

Answer
VerifiedVerified
536.1k+ views
Hint: First, take the natural logarithm of both sides of the equation to remove the variable from the exponent. Next, expand $\ln \left( {{{0.5}^x}} \right)$ and $\ln \left( {{{16}^2}} \right)$ by moving $x$ and $2$ respectively outside the logarithm. Next, write $0.5 = \dfrac{1}{2}$ and $16 = {2^4}$ in the equation. Next, expand $\ln \left( {{2^4}} \right)$ by moving $4$ outside the logarithm. Next, use the division property of natural logarithm by putting $m = 1$ and $n = 2$ in the equation. Next, put the natural logarithm of $1$ equal $0$. Next, divide each term by $\ln \left( 2 \right)$ and simplify. For this, divide each term in $x\ln \left( 2 \right) = 8\ln \left( 2 \right)$ by $\ln \left( 2 \right)$ and cancel the common factor of $\ln \left( 2 \right)$. Then, we will get the solution of the given equation.

Formula used:
$\ln \left( {{a^m}} \right) = m\ln \left( a \right)$
$\ln \left( {\dfrac{m}{n}} \right) = \ln \left( m \right) - \ln \left( n \right)$
$\ln \left( 1 \right) = 0$

Complete step by step answer:
Given equation: ${0.5^x} = {16^2}$
We have to find all possible values of $x$ satisfying given equation.
First, take the natural logarithm of both sides of the equation to remove the variable from the exponent.
$\ln \left( {{{0.5}^x}} \right) = \ln \left( {{{16}^2}} \right)$
Now, expand $\ln \left( {{{0.5}^x}} \right)$ and $\ln \left( {{{16}^2}} \right)$ by moving $x$ and $2$ respectively outside the logarithm as $\ln \left( {{a^m}} \right) = m\ln \left( a \right)$.
$ \Rightarrow x\ln \left( {0.5} \right) = 2\ln \left( {16} \right)$
Now, write $0.5 = \dfrac{1}{2}$ and $16 = {2^4}$ in the above equation.
$ \Rightarrow x\ln \left( {\dfrac{1}{2}} \right) = 2\ln \left( {{2^4}} \right)$
Now, expand $\ln \left( {{2^4}} \right)$ by moving $4$ outside the logarithm as $\ln \left( {{a^m}} \right) = m\ln \left( a \right)$.
$ \Rightarrow x\ln \left( {\dfrac{1}{2}} \right) = 2 \times 4 \times \ln \left( 2 \right)$
Multiply $2$ and $4$, we get
$ \Rightarrow x\ln \left( {\dfrac{1}{2}} \right) = 8\ln \left( 2 \right)$
Now, use property $\ln \left( {\dfrac{m}{n}} \right) = \ln \left( m \right) - \ln \left( n \right)$ by putting $m = 1$ and $n = 2$ in above equation.
$ \Rightarrow x\left[ {\ln \left( 1 \right) - \ln \left( 2 \right)} \right] = 8\ln \left( 2 \right)$
We know that the natural logarithm of $1$ is $0$, i.e., $\ln \left( 1 \right) = 0$.
$ \Rightarrow - x\ln \left( 2 \right) = 8\ln \left( 2 \right)$
Now, divide each term by $ - \ln \left( 2 \right)$ and simplify.
For this, divide each term in $ - x\ln \left( 2 \right) = 8\ln \left( 2 \right)$ by $ - \ln \left( 2 \right)$.
$ \Rightarrow \dfrac{{x\ln \left( 2 \right)}}{{\ln \left( 2 \right)}} = - \dfrac{{8\ln \left( 2 \right)}}{{\ln \left( 2 \right)}}$
Cancel the common factor of $\ln \left( 2 \right)$.
$ \Rightarrow x = - 8$

Hence, $x = - 8$ is the solution of ${0.5^x} = {16^2}$.

Note: In above question, we can find the solutions of given equation by plotting the equation, ${0.5^x} = {16^2}$ on graph paper and determine all its solutions.
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From the graph paper, we can see that $x = - 8$ is the solution of ${0.5^x} = {16^2}$.
Final solution: Hence, $x = - 8$ is the solution of ${0.5^x} = {16^2}$.