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Question

Given that `10^0.48  = x,  10^0.70 = y  and x^z = y^2 `,  then the value of  z is close to :


Options:
A .  1.45
B .  1.88
C .  2.9
D .  3.7
Answer: Option C

`x^z = y^2   hArr  (10^0.48)^z =  (10^0.70)^2     hArr  10^(0.48z)   =  10^(2 xx -.70)  = 10^1.40`

         `hArr`  0.48z =  1.40    ` hArr    z = 140/48          =  35/12 `    = 2.9 (approx.)




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