If $log 2 = a$ $log 2=a$ and $log 3 = b$ $log 3 = b$ , evaluate $log (\sqrt{60} \sqrt{2})$ $log(sqrt60sqrt2)$ ?

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Answer 1 · 2018-06-13T19:20:40

$\frac{3}{2} \cdot a + \frac{1}{2} \cdot b + \frac{1}{2} log 5$ $3/2*a+1/2*b+1/2log5$ .

Using the Usual Rules of $log$ $log$ ,

$log (\sqrt{60} \sqrt{2}) = log \sqrt{60} + log \sqrt{2}$ $log(sqrt60sqrt2)=logsqrt60+logsqrt2$

$= \frac{1}{2} log 60 + \frac{1}{2} log 2$ $=1/2log60+1/2log2$ ,

$= \frac{1}{2} log (2^{2} \cdot 3 \cdot 5) + \frac{1}{2} log 2$ $=1/2log(2^2*3*5)+1/2log2$ ,

$= \frac{1}{2} [{log 2}^{2} + log 3 + log 5] + \frac{1}{2} log 2$ $=1/2[log2^2+log3+log5]+1/2log2$ ,

$= \frac{1}{2} [2 log 2 + log 3 + log 5] + \frac{1}{2} log 2$ $=1/2[2log2+log3+log5]+1/2log2$ ,

$= (log 2 + \frac{1}{2} log 2) + \frac{1}{2} log 3 + \frac{1}{2} log 5$ $=(log2+1/2log2)+1/2log3+1/2log5$ ,

$= \frac{3}{2} log 2 + \frac{1}{2} log 3 + \frac{1}{2} log 5$ $=3/2log2+1/2log3+1/2log5$ ,

$\Rightarrow log (\sqrt{60} \sqrt{2}) = \frac{3}{2} \cdot a + \frac{1}{2} \cdot b + \frac{1}{2} log 5$ $rArr log(sqrt60sqrt2)=3/2*a+1/2*b+1/2log5$ .

If log2=alog 2=a and log3=blog 3 = b, evaluate log(√60√2)log(sqrt60sqrt2)?