The logarithm of 32sqrt[5]{4} to the base 2sq | vedclass

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Question

(A) Let $x$ be the required logarithm. By the definition of logarithm,we have: $(2\sqrt{2})^x = 32\sqrt[5]{4}$ Expressing both sides as powers of $2$:

Vedclass · Accepted Answer

$3.6$

Vedclass · Answer

(A) Let $x$ be the required logarithm. By the definition of logarithm,we have: $(2\sqrt{2})^x = 32\sqrt[5]{4}$ Expressing both sides as powers of $2$: $(2^1 \cdot 2^{1/2})^x = 2^5 \cdot (2^2)^{1/5}$ $(2^{3/2})^x = 2^5 \cdot 2^{2/5}$ $2^{3x/2} = 2^{5 + 2/5}$ $2^{3x/2} = 2^{27/5}$ Equating the exponents: $\frac{3x}{2} = \frac{27}{5}$ $x = \frac{27}{5} \cdot \frac{2}{3}$ $x = \frac{9 \cdot 2}{5} = \frac{18}{5} = 3.6$

The logarithm of $32\sqrt[5]{4}$ to the base $2\sqrt{2}$ is:

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