Question

Find the value of $(\frac{a}{b}{)}^3+(\frac{b}{a}{)}^3$ if $log(a^3-b^3)-log3=\frac{3}{2}(loga+logb)$.

Answer 1

log A + log B = log AB
log A - log B = $log\frac{A}{B}$
$log(a^3-b^3)-log3=\frac{3}{2}(loga+logb)$
$Property\Rightarrow log{x}^a=alogx$
$log\frac{a^3-b^3}{3}=\frac{3}{2}logab\Rightarrow log\frac{a^3-b^3}{3}=log(ab{)}^{\frac{3}{2}}$
$\therefore \frac{a^3-b^3}{3}=(ab{)}^{\frac{3}{2}}Squarebothside$
$\frac{(a^3-b^3{)}^2}{9}=(ab{)}^3$
$a^6+b^6-2a^3{b}^3=9a^3{b}^3$
$a^6+b^6=11a^3{b}^3$
$=\frac{a^3}{b^3}+\frac{b^3}{a^3}=\frac{a^6+b^6}{a^3{b}^3}=11$