log-3-12-log-36-3-log-3-4-log-108-3-x-x-

Question Number 168278 by Florian last updated on 07/Apr/22

\frac{{l o g}_{3} (12)}{{l o g}_{36} (3)} - \frac{{l o g}_{3} (4)}{{l o g}_{108} (3)} = x

x =

Commented by benhamimed last updated on 07/Apr/22

\frac{\frac{\ln 12}{\ln 3}}{\frac{\ln 3}{\ln 36}} - \frac{\frac{\ln 4}{\ln 3}}{\frac{\ln 3}{\ln 108}} = x

\frac{\ln 12 \times \ln 36 - \ln 4 \times \ln 108}{\ln^{2} 3} = x

\frac{(2 \ln 2 + \ln 3) 2 (\ln 2 + \ln 3) - 2 \ln 2 (2 \ln 2 + 3 \ln 3)}{\ln^{2} 3} = x

\frac{2 (\ln 2 + \ln 3) \ln 3 - 2 \ln 2 \ln 3}{\ln \overset{2}{} 3} = x

\frac{2 \ln 3}{\ln 3} = x

x = 2

Commented by Florian last updated on 07/Apr/22

V e r y G o o d!

Answered by greogoury55 last updated on 07/Apr/22

\Rightarrow x = \log_{3} (9 \times 4) . \log_{3} (3 \times 4) - \log_{3} (4) . \log_{3} (27 \times 4)

\Rightarrow x = (2 + \log_{3} (4)) (1 + \log_{3} (4)) - \log_{3} (4) (3 + \log_{3} (4))

l e t \log_{3} (4) = d

\Rightarrow x = (2 + d) (1 + d) - d (3 + d)

\Rightarrow x = 2 + 3 d + d^{2} - 3 d - d^{2}

\Rightarrow x = 2