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Question Number 106024 by john santu last updated on 02/Aug/20

log _4 (5x−6).log _x (256)=8

$$\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{5x}−\mathrm{6}\right).\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{256}\right)=\mathrm{8} \\ $$

Commented by Dwaipayan Shikari last updated on 02/Aug/20

{2,3}

$$\left\{\mathrm{2},\mathrm{3}\right\} \\ $$

Answered by Dwaipayan Shikari last updated on 02/Aug/20

((log(5x−6).log(256))/(log(4).log(x)))=8 ((log(5x−6).8log(2))/(2log(2)log(x)))=8 ((log(5x−6))/(log(x)))=2 5x−6=e^(2log(x)) 5x−6=x^2 x^2 −5x+6=0 { ((x=2)),((x=3)) :}

$$\frac{{log}\left(\mathrm{5}{x}−\mathrm{6}\right).{log}\left(\mathrm{256}\right)}{{log}\left(\mathrm{4}\right).{log}\left({x}\right)}=\mathrm{8} \\ $$$$\frac{{log}\left(\mathrm{5}{x}−\mathrm{6}\right).\mathrm{8}{log}\left(\mathrm{2}\right)}{\mathrm{2}{log}\left(\mathrm{2}\right){log}\left({x}\right)}=\mathrm{8} \\ $$$$\frac{{log}\left(\mathrm{5}{x}−\mathrm{6}\right)}{{log}\left({x}\right)}=\mathrm{2} \\ $$$$\mathrm{5}{x}−\mathrm{6}={e}^{\mathrm{2}{log}\left({x}\right)} \\ $$$$\mathrm{5}{x}−\mathrm{6}={x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}=\mathrm{0} \\ $$$$\begin{cases}{{x}=\mathrm{2}}\\{{x}=\mathrm{3}}\end{cases} \\ $$

Answered by bemath last updated on 02/Aug/20

(1/2)×8 log _2 (5x−6).log _x (2)=8 log _2 (5x−6)=2.log _2 (x) x^2 −5x+6=0 x = 2 & x = 3

$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{5}{x}−\mathrm{6}\right).\mathrm{log}\:_{{x}} \left(\mathrm{2}\right)=\mathrm{8} \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{5}{x}−\mathrm{6}\right)=\mathrm{2}.\mathrm{log}\:_{\mathrm{2}} \left({x}\right) \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}=\mathrm{0} \\ $$$${x}\:=\:\mathrm{2}\:\&\:{x}\:=\:\mathrm{3}\: \\ $$

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