Answer :
Sure! Let's solve the equation step by step:
The given equation is:
[tex]\[ 11 \cdot 2 \sqrt[5]{x} + 7 = 15 \][/tex]
Step 1: Isolate the radical expression.
First, we want to get the term containing the [tex]\( x \)[/tex] by itself. We can start by subtracting 7 from both sides:
[tex]\[ 11 \cdot 2 \sqrt[5]{x} = 15 - 7 \][/tex]
Simplify the right side:
[tex]\[ 11 \cdot 2 \sqrt[5]{x} = 8 \][/tex]
Step 2: Simplify the equation.
Next, simplify the equation by dividing both sides by 22 (because [tex]\( 11 \times 2 = 22 \)[/tex]):
[tex]\[ \sqrt[5]{x} = \frac{8}{22} \][/tex]
Simplify the fraction:
[tex]\[ \sqrt[5]{x} = \frac{4}{11} \][/tex]
Step 3: Solve for [tex]\( x \)[/tex].
Now, remove the fifth root by raising both sides of the equation to the power of 5:
[tex]\[ x = \left(\frac{4}{11}\right)^5 \][/tex]
Final Result:
Calculating [tex]\(\left(\frac{4}{11}\right)^5\)[/tex] gives us the decimal value:
[tex]\[ x \approx 0.0063582343481257495 \][/tex]
So, the solution to the equation is approximately [tex]\( x \approx 0.00636 \)[/tex].
The given equation is:
[tex]\[ 11 \cdot 2 \sqrt[5]{x} + 7 = 15 \][/tex]
Step 1: Isolate the radical expression.
First, we want to get the term containing the [tex]\( x \)[/tex] by itself. We can start by subtracting 7 from both sides:
[tex]\[ 11 \cdot 2 \sqrt[5]{x} = 15 - 7 \][/tex]
Simplify the right side:
[tex]\[ 11 \cdot 2 \sqrt[5]{x} = 8 \][/tex]
Step 2: Simplify the equation.
Next, simplify the equation by dividing both sides by 22 (because [tex]\( 11 \times 2 = 22 \)[/tex]):
[tex]\[ \sqrt[5]{x} = \frac{8}{22} \][/tex]
Simplify the fraction:
[tex]\[ \sqrt[5]{x} = \frac{4}{11} \][/tex]
Step 3: Solve for [tex]\( x \)[/tex].
Now, remove the fifth root by raising both sides of the equation to the power of 5:
[tex]\[ x = \left(\frac{4}{11}\right)^5 \][/tex]
Final Result:
Calculating [tex]\(\left(\frac{4}{11}\right)^5\)[/tex] gives us the decimal value:
[tex]\[ x \approx 0.0063582343481257495 \][/tex]
So, the solution to the equation is approximately [tex]\( x \approx 0.00636 \)[/tex].