Question

Find the value of $x$.

$$\begin{gathered} a){\log }_x\left(64\right)=2 \\ b){\log }_5\left(x\right)=3 \\ c){\log }_{\frac{1}{2}}\left(\frac{1}{4}\right)=x \end{gathered}$$

Short answer

Part (a) The value of $x$is$8$.

Part (b) The value of $x$is$125$.

Part (c) The value of $x$is$2$.

Step-by-step solution

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a){\log }_x\left(64\right)=2 \\ b){\log }_5\left(x\right)=3 \\ c){\log }_{\frac{1}{2}}\left(\frac{1}{4}\right)=x \end{gathered}$$

Step 2. Solving part (a).

Here, we will convert the given expression to the exponential form and solve the quadratic equation.

$$\begin{gathered} {\log }_x\left(64\right)=2 \\ {x}^2=64 \\ x=8,-8 \end{gathered}$$

The base of a logarithmic function cannot be negative hence the value of $x$is $8$.

Step 3. Solving part (b).

Now, we will convert the given expression to exponential form and simplify.

$$\begin{gathered} {\log }_5\left(x\right)=3 \\ x=5^3 \\ x=125 \end{gathered}$$

Therefore, the value of$x$is$125$.

Step 4. Solving part (c).

Now, we will convert the given expression to exponential form and simplify.

$$\begin{gathered} {\log }_{\frac{1}{2}}\left(\frac{1}{4}\right)=x \\ {\left(\frac{1}{2}\right)}^x=\frac{1}{4} \\ {\left(\frac{1}{2}\right)}^x={\left(\frac{1}{2}\right)}^2 \\ x=2 \end{gathered}$$

Therefore, the value of$x$is$2$.