Question

Give that ${10}^{0.48}=x$ and ${10}^{0.70}=y$ and $x^z=y^2$ then find the approximate value of z ?

• A. 2.9 approx
• B. 6.4 approx
• C. 3.0 approx
• D. 1.01 approx

Answer 1

The correct option is A 2.9 approx

Given $x^z=y^2\Rightarrow ({10}^{0.48}{)}^z=({10}^{0.70}{)}^2$
${10}^{0.48z}={10}^{1.40}\Rightarrow 0.48z=1.40\Rightarrow z=\frac{140}{48}=2.9\left(approx\right)$