If x^ -2 =64, then x^ 1 3 +x^0= 1. 2 2. 3 3. 3 2 4. 2 3

Question

If $\text{x}^{-2}=64,$ then $\text{x}^{\frac{1}{3}}+\text{x}^0=$

$2$
$3$
$\frac{3}{2}$
$\frac{2}{3}$

✓

Answer

$\frac{3}{2}$

Solution:
We have to find the value of $\text{x}^{\frac{1}{3}}+\text{x}^0$ if $\text{x}^{-2}=64$
Consider,
$\text{x}^{-2}=2^6$
$\frac{1}{\text{x}^2}=2^6$
Multiply $\frac{1}{2}$ on both sides of powers we get
$\frac{1}{\text{x}^{2\times\frac{1}{2}}}=2^{6\times\frac{1}{6}}$
$\frac{1}{\text{x}}=2^3$
$\frac{1}{\text{x}}=\frac{8}{1}$
By taking reciprocal on both sides we get,
$\frac{1}{8}=\text{x}$
Substituting $\frac{1}{8}$ in $\text{x}^{\frac{1}{3}}+\text{x}^0$ we get
$=\Big(\frac{1}{8}\Big)^{\frac{1}{3}}+\Big(\frac{1}{8}\Big)^0$
$=\Big(\frac{1}{2^2}\Big)^{\frac{1}{3}}+\Big(\frac{1}{8}\Big)^0$
$=\frac{1}{2^{3\times\frac{1}{3}}}+1$
$=\frac{1}{2^1}+1$
$=\frac{1}{2}+1$
By taking least common multiply we get
$=\frac{1}{2}+\frac{1\times2}{1\times2}$
$=\frac{1}{2}+\frac{2}{2}$
$=\frac{1+2}{2}$
$=\frac{3}{2}$
Hence the correct choice is c.

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