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log-2-3-x-log-3-5-y-lg6-




Question Number 144120 by mathdanisur last updated on 21/Jun/21
log_2 3 = x , log_3 5 = y , lg6 = ?
$${log}_{\mathrm{2}} \mathrm{3}\:=\:{x}\:,\:{log}_{\mathrm{3}} \mathrm{5}\:=\:{y}\:,\:{lg}\mathrm{6}\:=\:? \\ $$
Answered by Ar Brandon last updated on 21/Jun/21
log_3 5=y⇒log_2 5=xy  log6=log(2×3)           =log2+log3           =((1+log_2 3)/(log_2 10))           =((1+log_2 3)/(1+log_2 5))           =((1+x)/(1+xy))
$$\mathrm{log}_{\mathrm{3}} \mathrm{5}=\mathrm{y}\Rightarrow\mathrm{log}_{\mathrm{2}} \mathrm{5}=\mathrm{xy} \\ $$$$\mathrm{log6}=\mathrm{log}\left(\mathrm{2}×\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{log2}+\mathrm{log3} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}+\mathrm{log}_{\mathrm{2}} \mathrm{3}}{\mathrm{log}_{\mathrm{2}} \mathrm{10}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}+\mathrm{log}_{\mathrm{2}} \mathrm{3}}{\mathrm{1}+\mathrm{log}_{\mathrm{2}} \mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}+\mathrm{xy}} \\ $$
Commented by mathdanisur last updated on 21/Jun/21
cool Sir thankyou
$${cool}\:{Sir}\:{thankyou} \\ $$
Answered by liberty last updated on 22/Jun/21
log _(10) (6)= log _(10) (2)+log _(10) (3)                    = (1/(1+log _2 (5)))  +((log _2 (3))/(1+log _2 (5)))                    = ((1+x)/(1+xy)) .
$$\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{6}\right)=\:\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2}\right)+\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{5}\right)}\:\:+\frac{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}\right)}{\mathrm{1}+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{5}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}+\mathrm{xy}}\:. \\ $$

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