If $b^{- \frac{1}{2}} = 4$ $b^(-1/2) = 4$ , what is the value of b?

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Answer 1 · 2015-09-04T13:29:42

You first rewrite the equation.

The $-$ $-$ in a power means it is 1 divided by.
$\frac{1}{2}$ $1/2$ as a power means $\sqrt{}$ $sqrt$

So $b^{- \frac{1}{2}}$ $b^(-1/2)$ means $\frac{1}{\sqrt{b}}$ $1/sqrtb$

$\frac{1}{\sqrt{b}} = 4 \to 4 \cdot \sqrt{b} = 1 \to \sqrt{b} = \frac{1}{4} \to$ $1/sqrtb=4->4*sqrtb=1->sqrtb=1/4->$

Square both sides:
${\sqrt{b}}^{2} = {(\frac{1}{4})}^{2} \to b = \frac{1}{16}$ $sqrtb^2=(1/4)^2->b=1/16$

Check your answer by doing it the other way around:

${(\frac{1}{16})}^{- 1} = \frac{1}{\frac{1}{16}} = 16 \to$ $(1/16)^-1=1/(1/16)=16->$
$16^{\frac{1}{2}} = \sqrt{16} = 4$ $16^(1/2)=sqrt16=4$

If b^(-1/2) = 4b−12=4b^(-1/2) = 4, what is the value of b?