Question

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

• A. 6
• B. 7
• C. 8
• D. 5

Answer 1

Answer: (B)

7

Answer 2

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