Question

If \(2^x\) = \(\sqrt[3]{64}\) ,then \(x^2 + x + 1\) = ?

• A. 6
• B. 7
• C. 8
• D. 5

Answer 1

The Correct Option is B

To solve the equation \(f\)⋅\(2^x\)=\(\sqrt[3]{64}\) ,we first need to find the value of \(\sqrt[3]{64}\)
\(\sqrt[3]{64}\) = 4 (since 4³=64)
Rearranging the equation:
\(f\)⋅\(2^x\)= 4
Assuming \(f\) is a constant that we will solve for later, we can divide both sides by \(f\):
⇒\(2^x=\frac{4}{f}\)
Take the logarithm: Taking the logarithm base 2 on both sides:
\(x=\log_2(\frac{4}{f})\)
Next, we need to find \(x^2+x+1\)
Calculate \(x\):
\(x=\log_2(2)^2-log_2(f)=2-log_2(f)\)
Substituting into \(x^2+x+1\)
\(x^2=(2-log_2(f))^2=4-4log_2(f)+(log_2(f))^2\)
Now add \(x\) and 1:
\(x^2+x+1=(4-\log_2(f)+(log_2(f))^2)+(2-log_2(f))+1\)
\(=4+2+1-4log_2(f)-log_2(f)+(log_2(f))^2\)
\(=7-5log_2(f)+(log_2(f))^2\)
Without knowing the exact value of \(f\), we cannot simplify further to arrive at one of the answer choices directly.
If you meant \(f\)=1, then:
\(\log_2(1)=0\)
\(x=2\)
\(x^2+x+1=2^2+2+1=4+2+1=7\)
So the correct option is (B):7