Log in with a correct answer

Given that $\begin{cases} \dfrac {\log 3}{\log 30} = a & \implies \log 3 = a \log 30 \\ \dfrac {\log 5}{\log 30} = b & \implies \log 5 = b \log 30 \end{cases}$

Now we have:

$\begin{aligned} \log 30 & = \log 2 + \log 3 + \log 5 \\ \implies \log 2 & = \log 30 - \log 3 - \log 5 \\ & = \log 30 - a\log 30 - b\log 30 \\ & = \log 30(1-a-b) \\ 3 \log 2 & = 3 \log 30(1-a-b) \\ \implies \frac {\log 8}{\log 30} & = \boxed{3(1-a-b)} \end{aligned}$

• On addition we get, $\frac{log15}{log30}$ = a+ b
  $\frac{log15}{log15*2}$ = a + b
  $\frac{log15}{log15 + log2}$ = a + b
  Taking reciprocal we get,
  $\frac{log2}{log15}$ = $\frac{1 - a - b}{a + b}$
  Substituting the value of a + b i.e. $\frac{log15}{log30}$ we get,
  $\frac{log2}{log15}$ = (1 - a - b) $\frac{log15}{log30}$
  On solving further we get,
  $\frac{log8}{log30}$ = 3(1 - a - b)
  
  
  

$\begin{aligned} \log_{30} 3 = a; & \ \log_{30} 5 = b \\ \\ \log_{30} 15 &= a + b \\ \\ \log_{30} 2 &= \log_{30} 30 - \log_{30} 15 \\ &= 1 - a - b \\ \\ \log_{30} 8 &= 3 \log_{30} 2 \\ &= 3 (1 - a - b) \end{aligned}$