Question

Given that ${10}^{0.48}=x,{10}^{0.70}=yand{x}^z=y^2,$then the value of z is close to :

• A. 1.45
• B. 1.88
• C. 2.9
• D. 3.7
• E. 1.35

Answer 1

The correct option is C

2.9

$\begin{array}{cc}{x}^2=y^2\iff {\left({10}^{0.48}\right)}^z={\left({10}^{0.70}\right)}^2\iff {10}^{\left(0.48z\right)}={10}^{\left(2\times 0.70\right)}={10}^{1.40}& \\ & \iff 0.48z=1.40\iff z=\frac{140}{48}=2.935\left(approx\right).\end{array}$