Question

Given that ${10}^{0.48}=xand{10}^{0.7}=y$ and $x^z=y^2$, find the value of z

• A. $2\frac{11}{12}$
• B. $\frac{4}{9}$
• C. $1\frac{1}{48}$
• D. $\frac{48}{49}$

Answer 1

The correct option is A $2\frac{11}{12}$

Given $x^z=y^2\Rightarrow ({10}^{0.48}{)}^z=({10}^{0.70}{)}^2$
${10}^{0.48z}={10}^{1.40}\Rightarrow 0.48z=1.40\Rightarrow z=\frac{140}{48}=\frac{35}{12}=2\frac{11}{12}$