Question

Find the value of $x$.

$$\begin{gathered} a){\log }_x\left(81\right)=2 \\ b){\log }_3\left(x\right)=5 \\ c){\log }_{\frac{1}{3}}\left(\frac{1}{27}\right)=x \end{gathered}$$

Short answer

Part (a) The value of $x$is$9$.

Part (b) The value of $x$is$243$.

Part (c) The value of$x$is$3$.

Step-by-step solution

Step 1. Given information.

The given expressions are,

$$\begin{gathered} a){\log }_x\left(81\right)=2 \\ b){\log }_3\left(x\right)=5 \\ c){\log }_{\frac{1}{3}}\left(\frac{1}{27}\right)=x \end{gathered}$$

Solving part (a).

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

$$\begin{gathered} {\log }_x\left(81\right)=2 \\ {x}^2=81 \\ x=9,-9 \end{gathered}$$

The base of a logarithmic function cannot be negative hence the value of$x$is$9$.

Step 3. Solving part (b).

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

$$\begin{gathered} {\log }_3\left(x\right)=5 \\ x=3^5 \\ x=243 \end{gathered}$$

Therefore the value of$x$is$243$.

Step 4. Solving part (c).

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

$$\begin{gathered} {\log }_{\frac{1}{3}}\left(\frac{1}{27}\right)=x \\ {\left(\frac{1}{3}\right)}^x=\frac{1}{27} \\ {\left(\frac{1}{3}\right)}^x={\left(\frac{1}{3}\right)}^3 \\ x=3 \end{gathered}$$

Therefore the value of$x$is$3$.