Given that $\displaystyle 2^a = 3^b = 18^c$, show that ab = c(b + 2a)
$\displaystyle 2^a=3^b$
$\displaystyle aln(2)=b ln(3) $
$\displaystyle ln(3)=\frac{a ln(2)}{b}$ .....(1)
$\displaystyle 3^b=18^c$
$\displaystyle 3^b=(2(3)^2)^c$
$\displaystyle b ln(3) = c ln(2 (3^2))$
$\displaystyle b ln(3) = c(ln(2) + 2ln(3) ) $ since $\displaystyle ln(AB) = ln(A) + ln(B) $
sub the value of ln(3) from (1)
$\displaystyle b\left(\frac{a ln(2)}{b}\right) = c(ln(2) + 2\left(\frac{a ln(2)}{b}\right) $
$\displaystyle b\left(a ln(2)\right) = cbln(2) + 2c\left(a ln(2)\right) $ by multiply with b
$\displaystyle b\left(a \right) = cb + 2c\left(a \right) $ division by ln(2)
$\displaystyle ba = c(b+2a) $
$\displaystyle \ln$ denotes the natural log, but for this any base of logarithm can be used. \log often denotes a logarithm to the base $\displaystyle 10$, but just as often denotes a natural logarithm or a logarithm to an unspecified base, so yes you can use $\displaystyle \log$.
(any base will do for this problem)
CB