Chapter 6.3, Problem 42E

To determine

Check the error and rewrite the expression in correct way.

Answer to Problem 42E

The correct solution is ${\log }_4{64}^x=3x$ .

Explanation of Solution

Given information:

The given logerithmic solution is ${\log }_4{64}^x=2+x$ .

Formula Used:

Logarithmic power rule is

If $a^b$ is an expression then logarithm form ${\log }_a{\left(a\right)}^b$ is equal to $b$ .

Calculation:

Consider the logarithmic expression is ${\log }_4{64}^x$ .

In the question the solution is given as,

  $\begin{array}{c}{\log }_4{64}^x={\log }_4\left(16\times 4^x\right)\\ ={\log }_4\left(4^2\times 4^x\right)\\ ={\log }_4{4}^{2+x}\\ =2+x\end{array}$

But it is wrong to break ${64}^x$ as $16\times 4^x$ , it should be equal to ${\left(4^3\right)}^x$ . Thus, the given solution is incorrect.

Now, find the solution in correct way.

Solve the expression ${\log }_4{64}^x$ . Use the logarithmic power rule.

  $\begin{array}{c}{\log }_4{64}^x={\log }_4{\left(4^3\right)}^x\\ ={\log }_4{4}^3{}^x\\ =3x\times {\log }_{4}4\\ =3x\end{array}$

Hence, the correct solution is ${\log }_4{64}^x=3x$ .