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

$$\frac{{log}_3\left(12\right)}{{log}_{36}\left(3\right)}-\frac{{log}_3\left(4\right)}{{log}_{108}\left(3\right)}=x$$$$x=$$

Commented by benhamimed last updated on 07/Apr/22

$$\frac{\frac{\ln 12}{\ln 3}}{\frac{\ln 3}{\ln 36}}-\frac{\frac{\ln 4}{\ln 3}}{\frac{\ln 3}{\ln 108}}=x$$$$\frac{\ln 12\times \ln 36-\ln 4\times \ln 108}{\ln {}^{2}3}=x$$$$\frac{(2\ln 2+\ln 3)2(\ln 2+\ln 3)-2\ln 2(2\ln 2+3\ln 3)}{\ln {}^{2}3}=x$$$$\frac{2(\ln 2+\ln 3)\ln 3-2\ln 2\ln 3}{\ln \stackrel{2}{}3}=x$$$$\frac{2\ln 3}{\ln 3}=x$$$$x=2$$

Commented by Florian last updated on 07/Apr/22

$$VeryGood!$$

Answered by greogoury55 last updated on 07/Apr/22

$$\Rightarrow x=\log {}_3\left(9\times 4\right).\log {}_3\left(3\times 4\right)-\log {}_3\left(4\right).\log {}_3\left(27\times 4\right)$$$$\Rightarrow x=(2+\log {}_3\left(4\right))(1+\log {}_3\left(4\right))-\log {}_3\left(4\right)(3+\log {}_3\left(4\right))$$$$let\log {}_3\left(4\right)=d$$$$\Rightarrow x=(2+d)(1+d)-d(3+d)$$$$\Rightarrow x=2+3d+d^2-3d-d^2$$$$\Rightarrow x=2$$

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