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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
