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

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Question

(A) Let the required logarithm be $x$. Then,by definition,$(2\sqrt{2})^x = 32\sqrt[5]{4}$.We can write the base as $(2 \cdot 2^{1/2})^x = (2^{3/2})^x

Vedclass · Accepted Answer

$3.6$

Vedclass · Answer

(A) Let the required logarithm be $x$. Then,by definition,$(2\sqrt{2})^x = 32\sqrt[5]{4}$.We can write the base as $(2 \cdot 2^{1/2})^x = (2^{3/2})^x = 2^{3x/2}$.We can write the number as $32 \cdot 4^{1/5} = 2^5 \cdot (2^2)^{1/5} = 2^5 \cdot 2^{2/5} = 2^{5 + 2/5} = 2^{27/5}$.Equating the exponents,we get $\frac{3x}{2} = \frac{27}{5}$.Solving for $x$,$x = \frac{27}{5} 	imes \frac{2}{3} = \frac{9}{5} 	imes 2 = \frac{18}{5} = 3.6$.

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

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