Question

Given that ${10}^{0.48}=x,{10}^{0.70}=y$ and $x^z=y^2$, then the value of $z$ is (approximately) _________.

• A. $1.45$
• B. $1.88$
• C. $2.9$
• D. $3.7$

Answer 1

The correct option is C $2.9$

Given that, $x^z=y^2$

Also ${10}^{0.48}=x,{10}^{0.70}=y$

Taking $\log$ on both sides with the given equation, we get, ------($\log$ with respect to base $10$)

$\log (x{)}^z=\log (y{)}^2$

Applying property $\log (a{)}^b=b.\log (a)$,we get

$z\log \left(x\right)=2\log \left(y\right)$

$z=\frac{2\log \left(y\right)}{\log \left(x\right)}$

$=\frac{2\log \left({10}^{0.7}\right)}{\log \left({10}^{0.48}\right)}$

$=\frac{2\times 0.7}{0.48}=2.91666$