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Question Number 22345 by ibraheem160 last updated on 16/Oct/17

6^(x+2) =2(3^x ), find x?

$$\mathrm{6}^{\mathrm{x}+\mathrm{2}} =\mathrm{2}\left(\mathrm{3}^{\mathrm{x}} \right),\:\mathrm{find}\:\mathrm{x}? \\ $$

Commented by JI Siam last updated on 21/Dec/17

Here 6^(x+1) =3^(x−1) so,likely x is not a normal number.

$$\mathrm{Here}\:\mathrm{6}^{\mathrm{x}+\mathrm{1}} =\mathrm{3}^{\mathrm{x}−\mathrm{1}} \:\mathrm{so},\mathrm{likely}\:\mathrm{x}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{normal}\:\mathrm{number}. \\ $$

Answered by $@ty@m last updated on 16/Oct/17

3^(x+2) .2^(x+2) =2.3^x (3^(x+2) /3^x )=(2/2^(x+2) ) 3^2 =(1/2^(x+1) ) 2^(x+1) =(1/3^2 ) log 2^(x+1) =log (1/3^2 ) (x+1)log 2=log 1−log 3^2 (x+1)log 2=−2log 3 x+1=((−2log 3)/(log 2)) x=((−2log 3)/(log 2))−1

$$\mathrm{3}^{{x}+\mathrm{2}} .\mathrm{2}^{{x}+\mathrm{2}} =\mathrm{2}.\mathrm{3}^{{x}} \\ $$$$\frac{\mathrm{3}^{{x}+\mathrm{2}} }{\mathrm{3}^{{x}} }=\frac{\mathrm{2}}{\mathrm{2}^{{x}+\mathrm{2}} } \\ $$$$\mathrm{3}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}^{{x}+\mathrm{1}} } \\ $$$$\mathrm{2}^{{x}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} } \\ $$$$\mathrm{log}\:\mathrm{2}^{{x}+\mathrm{1}} =\mathrm{log}\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} } \\ $$$$\left({x}+\mathrm{1}\right)\mathrm{log}\:\mathrm{2}=\mathrm{log}\:\mathrm{1}−\mathrm{log}\:\mathrm{3}^{\mathrm{2}} \\ $$$$\left({x}+\mathrm{1}\right)\mathrm{log}\:\mathrm{2}=−\mathrm{2log}\:\mathrm{3} \\ $$$${x}+\mathrm{1}=\frac{−\mathrm{2log}\:\mathrm{3}}{\mathrm{log}\:\mathrm{2}} \\ $$$${x}=\frac{−\mathrm{2log}\:\mathrm{3}}{\mathrm{log}\:\mathrm{2}}−\mathrm{1} \\ $$

Commented by $@ty@m last updated on 16/Oct/17

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