Question

Show that : $7\log \left(\frac{16}{15}\right)+5\log \left(\frac{25}{24}\right)+3\log \left(\frac{81}{80}\right)=\log 2$.

Answer 1

$L.H.S$

$7log\left(\frac{16}{15}\right)+5log\left(\frac{25}{24}\right)+3log\left(\frac{81}{80}\right)$

$=7log16-7log15+5log25-5log24+3log81-3log80(log\frac{b}{a}=logb-loga)$

$=7log{2}^4-7log3.5+5log{5}^2-5log{2}^3.3+3log{3}^4-3log{2}^4.5$

$=28log2-7log3-7log5+10log5-5log{2}^3-5log3+12log3-3log{2}^4-3log5$

$\left[\begin{array}{c}loga\cdot b=loga+logb\\ log{x}^n=nlogx\end{array}\right]$

$=28log2-7log3+3log5-15log2+7log3-12log2-3log5$

$=28log2-15log2-12log2$

$=28log2-27log2$

$=log2$

$=R.H.S.$