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log-3-x-1-log-4-x-8-x-




Question Number 145067 by imjagoll last updated on 02/Jul/21
 log _3 (x+1) =log _4 (x+8)   x=?
$$\:\mathrm{log}\:_{\mathrm{3}} \left({x}+\mathrm{1}\right)\:=\mathrm{log}\:_{\mathrm{4}} \left({x}+\mathrm{8}\right) \\ $$$$\:{x}=? \\ $$
Answered by bobhans last updated on 02/Jul/21
 x=8 ⇒log _3 (8+1)=log _4 (8+8)  ⇒ 2 = 2
$$\:\mathrm{x}=\mathrm{8}\:\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{8}+\mathrm{1}\right)=\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{8}+\mathrm{8}\right) \\ $$$$\Rightarrow\:\mathrm{2}\:=\:\mathrm{2}\: \\ $$
Commented by bobhans last updated on 02/Jul/21
let log _3 (x+1)=k ⇒x+1=3^k    and log _4 (x+8)=k⇒x+8=4^k   ⇒4^k −3^k  = 7   k=0⇒1−1≠7  k=2⇒16−9=7 then x=3^2 −1 = 8 or  x=4^k −8=16−8=8
$$\mathrm{let}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}+\mathrm{1}\right)=\mathrm{k}\:\Rightarrow\mathrm{x}+\mathrm{1}=\mathrm{3}^{\mathrm{k}} \\ $$$$\:\mathrm{and}\:\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{x}+\mathrm{8}\right)=\mathrm{k}\Rightarrow\mathrm{x}+\mathrm{8}=\mathrm{4}^{\mathrm{k}} \\ $$$$\Rightarrow\mathrm{4}^{\mathrm{k}} −\mathrm{3}^{\mathrm{k}} \:=\:\mathrm{7}\: \\ $$$$\mathrm{k}=\mathrm{0}\Rightarrow\mathrm{1}−\mathrm{1}\neq\mathrm{7} \\ $$$$\mathrm{k}=\mathrm{2}\Rightarrow\mathrm{16}−\mathrm{9}=\mathrm{7}\:\mathrm{then}\:\mathrm{x}=\mathrm{3}^{\mathrm{2}} −\mathrm{1}\:=\:\mathrm{8}\:\mathrm{or} \\ $$$$\mathrm{x}=\mathrm{4}^{\mathrm{k}} −\mathrm{8}=\mathrm{16}−\mathrm{8}=\mathrm{8} \\ $$

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