Q. 8.120 - Intermediate Algebra

Question

Solve: ${\left(4x-8\right)}^{\frac{1}{4}}+5=7$.

Step-by-Step Solution

Answer

The solution is, $x=6$.

Step 1 Given equation is,

${\left(4x-8\right)}^{\frac{1}{4}}+5=7$.

The objective is, we need to find the solution for the given equation.

Step 2 To isolate the radical, subtract 5 on both sides.

$$\begin{gathered} {\left(4x-8\right)}^{\frac{1}{4}}+5-5=7-5 \\ {\left(4x-8\right)}^{\frac{1}{4}}=2 \end{gathered}$$

Raise each side of the equation to fourth power.

${\left({\left(4x-8\right)}^{\frac{1}{4}}\right)}^4={\left(2\right)}^4$

Simplify it.

$$\begin{gathered} {\left({\left(4x-8\right)}^{\frac{1}{4}}\right)}^4={\left(2\right)}^4 \\ 4x-8=16 \\ 4x=16+8 \\ 4x=24 \\ x=\frac{24}{4} \\ x=6 \end{gathered}$$

Step 3 Let us check the answer.

Substitute $x=6$ into the equation

$$\begin{gathered} {\left(4x-8\right)}^{\frac{1}{4}}+5=7 \\ {\left(4\left(6\right)-8\right)}^{\frac{1}{4}}+5=7 \\ {\left(24-8\right)}^{\frac{1}{4}}+5=7 \\ {\left(16\right)}^{\frac{1}{4}}+5=7 \\ {\left(2^4\right)}^{\frac{1}{4}}+5=7 \\ 2+5=7 \\ 7=7 \end{gathered}$$

which is true.