Basic of Logarithms
easy

Logarithm of $32\root 5 \of 4 $ to the base $2\sqrt 2 $ is

A

$3.6$

B

$5$

C

$5.6$

D

None of these

Solution

(a) Let $x$ be the required logarithm , then by definition

${(2\sqrt 2 )^x} = 32\root 5 \of 4 $

==> ${({2.2^{1/2}})^x} = {2^5}{.2^{2/5}}$;  ${2^{{{3x} \over 2}}} = {2^{5 + {2 \over 5}}}$

Here, by equating the indices, ${3 \over 2}x = {{27} \over 5}$

$\therefore x = \frac{{18}}{5} = 3.6$
.

Standard 11
Mathematics

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