Given 10^{0.48} = x, 10^{0.70} = y, and x^z = y^2, find z (approximate).

Given data

• x = 10^{0.48}, y = 10^{0.70}, and x^z = y^2.

Concept / Approach

• Use logarithms/exponent rules: (10^{a})^{b} = 10^{ab}.

Step-by-step calculation

x^z = (10^{0.48})^{z} = 10^{0.48z}y^2 = (10^{0.70})^{2} = 10^{1.40}Equate exponents: 0.48z = 1.40 ⇒ z = 1.40 / 0.48 = 140/48 = 35/12 ≈ 2.92.

Verification

(10^{0.48})^{35/12} = 10^{(0.48×35/12)} = 10^{1.40} = (10^{0.70})^2.

Common pitfalls

• Adding 0.48 and 0.70 or squaring incorrectly; remember to equate exponents.

Final Answer

2.92 (approximately).