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Question Number 99257 by peter frank last updated on 19/Jun/20

Answered by Ar Brandon last updated on 19/Jun/20

1a\log_(ab) x=((log_x x)/(log_x ab))=(1/(log_x a+log_x b))=(1/(((log_a a)/(log_a x))+((log_b b)/(log_b x)))) =(1/((1/(log_a x))+(1/(log_b x))))=(1/({((log_b x+log_a x)/((log_a x)(log_b x)))})) =(((log_a x)(log_b x))/(log_a x+log_b x))

$$\mathrm{1a}\backslash\mathrm{log}_{\mathrm{ab}} \mathrm{x}=\frac{\mathrm{log}_{\mathrm{x}} \mathrm{x}}{\mathrm{log}_{\mathrm{x}} \mathrm{ab}}=\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{a}+\mathrm{log}_{\mathrm{x}} \mathrm{b}}=\frac{\mathrm{1}}{\frac{\mathrm{log}_{\mathrm{a}} \mathrm{a}}{\mathrm{log}_{\mathrm{a}} \mathrm{x}}+\frac{\mathrm{log}_{\mathrm{b}} \mathrm{b}}{\mathrm{log}_{\mathrm{b}} \mathrm{x}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{a}} \mathrm{x}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{b}} \mathrm{x}}}=\frac{\mathrm{1}}{\left\{\frac{\mathrm{log}_{\mathrm{b}} \mathrm{x}+\mathrm{log}_{\mathrm{a}} \mathrm{x}}{\left(\mathrm{log}_{\mathrm{a}} \mathrm{x}\right)\left(\mathrm{log}_{\mathrm{b}} \mathrm{x}\right)}\right\}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{log}_{\mathrm{a}} \mathrm{x}\right)\left(\mathrm{log}_{\mathrm{b}} \mathrm{x}\right)}{\mathrm{log}_{\mathrm{a}} \mathrm{x}+\mathrm{log}_{\mathrm{b}} \mathrm{x}} \\ $$

Answered by mahdi last updated on 20/Jun/20

1)log_(ab) x=(1/(log_x ab))=(1/(1og_x a+log_x b))= (1/((1/(log_a x))+(1/(log_b x))))=((log_b x.log_a x)/(1og_b x+log_a x)) (not=((log_b x−log_a x)/(log_b x+log_a x))) 2)2log(a+b)=log(a^2 +b^2 +2ab)= log(a^2 ×(1+(b^2 /a^2 )+((2b)/a)))=2loga+log(1+(b^2 /a^2 )+((2b)/a)) (not log_c ) 3)=log_2 (((sinx)/(cosx(1−tanx)(1+tanx))))= log_2 (((sinx)/(cosx(1−tan^2 x))))=log_2 (((sinx)/(cosx−tanxsinx))) log_2 (((sinxcosx)/(cos^2 x−sin^2 x)))=log_2 ((((1/2)×sin(2x))/(cos(2x))))= log_2 (tan(2x))−1

$$\left.\mathrm{1}\right)\mathrm{log}_{\mathrm{ab}} \mathrm{x}=\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{ab}}=\frac{\mathrm{1}}{\mathrm{1og}_{\mathrm{x}} \mathrm{a}+\mathrm{log}_{\mathrm{x}} \mathrm{b}}= \\ $$$$\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{a}} \mathrm{x}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{b}} \mathrm{x}}}=\frac{\mathrm{log}_{\mathrm{b}} \mathrm{x}.\mathrm{log}_{\mathrm{a}} \mathrm{x}}{\mathrm{1og}_{\mathrm{b}} \mathrm{x}+\mathrm{log}_{\mathrm{a}} \mathrm{x}}\:\:\left(\mathrm{not}=\frac{\mathrm{log}_{\mathrm{b}} \mathrm{x}−\mathrm{log}_{\mathrm{a}} \mathrm{x}}{\mathrm{log}_{\mathrm{b}} \mathrm{x}+\mathrm{log}_{\mathrm{a}} \mathrm{x}}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{2log}\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{log}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{2ab}\right)= \\ $$$$\mathrm{log}\left(\mathrm{a}^{\mathrm{2}} ×\left(\mathrm{1}+\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{2b}}{\mathrm{a}}\right)\right)=\mathrm{2loga}+\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{2b}}{\mathrm{a}}\right)\:\:\left(\mathrm{not}\:\mathrm{log}_{\mathrm{c}} \right) \\ $$$$\left.\mathrm{3}\right)=\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{sinx}}{\mathrm{cosx}\left(\mathrm{1}−\mathrm{tanx}\right)\left(\mathrm{1}+\mathrm{tanx}\right)}\right)= \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{sinx}}{\mathrm{cosx}\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)}\right)=\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{sinx}}{\mathrm{cosx}−\mathrm{tanxsinx}}\right) \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{sinxcosx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\right)=\mathrm{log}_{\mathrm{2}} \left(\frac{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{sin}\left(\mathrm{2x}\right)}{\mathrm{cos}\left(\mathrm{2x}\right)}\right)= \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\mathrm{tan}\left(\mathrm{2x}\right)\right)−\mathrm{1} \\ $$

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