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Question Number 131457 by pete last updated on 04/Feb/21
Find x in the equation 3^(x+1) = 4^(x−1)
$$\mathrm{Find}\:{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{3}^{{x}+\mathrm{1}} \:=\:\mathrm{4}^{{x}−\mathrm{1}} \\ $$
Answered by bramlexs22 last updated on 05/Feb/21
⇔ (x+1) ln 3 = (x−1)ln 4 ⇔ x(ln 3−ln 4) = −ln 4−ln 3 ⇔ x(ln 4−ln 3)=ln 4+ln 3 ⇔ x = ((ln 12)/(ln ((4/3)))) = log _(((4/3))) (12)
$$\Leftrightarrow\:\left({x}+\mathrm{1}\right)\:\mathrm{ln}\:\mathrm{3}\:=\:\left({x}−\mathrm{1}\right)\mathrm{ln}\:\mathrm{4} \\ $$$$\Leftrightarrow\:{x}\left(\mathrm{ln}\:\mathrm{3}−\mathrm{ln}\:\mathrm{4}\right)\:=\:−\mathrm{ln}\:\mathrm{4}−\mathrm{ln}\:\mathrm{3} \\ $$$$\Leftrightarrow\:{x}\left(\mathrm{ln}\:\mathrm{4}−\mathrm{ln}\:\mathrm{3}\right)=\mathrm{ln}\:\mathrm{4}+\mathrm{ln}\:\mathrm{3} \\ $$$$\Leftrightarrow\:{x}\:=\:\frac{\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)}\:=\:\mathrm{log}\:_{\left(\frac{\mathrm{4}}{\mathrm{3}}\right)} \left(\mathrm{12}\right) \\ $$

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