$32\root 5 \of 4 $ to the base $2\sqrt 2 = . . . .$
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a
(a) Let \(x\) be the required logarithm , then by definition
(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, \({3 \over 2}x = {{27} \over 5}\)
\(\therefore x = \frac{{18}}{5} = 3.6\)
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