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Question Number 176189 by Linton last updated on 14/Sep/22

solve for x  (1+(1/x))^(x+1) =(1+(1/6))^6   show working

$${solve}\:{for}\:{x} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}+\mathrm{1}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{6}} \\ $$$${show}\:{working} \\ $$

Answered by BaliramKumar last updated on 14/Sep/22

(((x+1)/x))^(x+1)  = (1+(1/6))^6   ((x/(x+1)))^(−(x+1))  = (1+(1/6))^6   (((x+1−1)/(x+1)))^(−(x+1))  = (1+(1/6))^6   (1−(1/(x+1)))^(−(x+1))  = (1+(1/6))^6   put      −(x+1) = y  (1+(1/y))^y  = (1+(1/6))^6   y = 6  −(x+1) = 6  x+1 = −6  x = −7 Answer

$$\left(\frac{{x}+\mathrm{1}}{{x}}\right)^{{x}+\mathrm{1}} \:=\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{6}} \\ $$$$\left(\frac{{x}}{{x}+\mathrm{1}}\right)^{−\left({x}+\mathrm{1}\right)} \:=\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{6}} \\ $$$$\left(\frac{{x}+\mathrm{1}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{−\left({x}+\mathrm{1}\right)} \:=\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{6}} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{−\left({x}+\mathrm{1}\right)} \:=\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{6}} \\ $$$${put}\:\:\:\:\:\:−\left({x}+\mathrm{1}\right)\:=\:{y} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{y}}\right)^{{y}} \:=\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{6}} \\ $$$${y}\:=\:\mathrm{6} \\ $$$$−\left({x}+\mathrm{1}\right)\:=\:\mathrm{6} \\ $$$${x}+\mathrm{1}\:=\:−\mathrm{6} \\ $$$${x}\:=\:−\mathrm{7}\:\mathrm{Answer} \\ $$$$ \\ $$

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