32root 5 of 4 to the base 2sqrt 2 = . . . . | vedclass

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Question

(a) Let $x$ be the required logarithm , then by definition

${(2\sqrt 2 )^x} = 32\root 5 \of 4 $ 

==> ${({2.2^{1/2}})^x} = {2^5}{.2^{2/5}

Vedclass · Accepted Answer

$3.6$

Vedclass · Answer

(a) Let $x$ be the required logarithm , then by definition

${(2\sqrt 2 )^x} = 32oot 5 \of 4 $ 

==> ${({2.2^{1/2}})^x} = {2^5}{.2^{2/5}}$;  ${2^{{{3x} \over 2}}} = {2^{5 + {2 \over 5}}}$

Here, by equating the indices, ${3 \over 2}x = {{27} \over 5}$

$	herefore x = \frac{{18}}{5} = 3.6$
.

$32\root 5 \of 4 $ to the base $2\sqrt 2 = . . . .$

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