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Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
Answer & Solution
Correct Answer:
Option
C
$$\eqalign{
& {x^z} = {y^2} \Leftrightarrow {10^{\left( {0.48z} \right)}} = {10^{2 \times 0.70}} = {10^{1.40}} \cr
& \Rightarrow 0.48z = 1.40 \cr
& \Rightarrow z = \frac{{140}}{{48}} = \frac{{35}}{{12}} = 2.9({\text{approx}}) \cr} $$
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