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Question Number 144408 by mathdanisur last updated on 25/Jun/21

lg(2)=x lg(7)=y lg_5 (9,8)=?

$$\boldsymbol{{lg}}\left(\mathrm{2}\right)=\boldsymbol{{x}} \\ $$$$\boldsymbol{{lg}}\left(\mathrm{7}\right)=\boldsymbol{{y}} \\ $$$$\boldsymbol{{lg}}_{\mathrm{5}} \left(\mathrm{9},\mathrm{8}\right)=? \\ $$

Answered by liberty last updated on 25/Jun/21

{ ((log _(10) (2)=x)),((log _(10) (7)=y)) :}⇒ log _5 (9.8)=? ⇔ log _5 (9.8)= ((log _(10) (98)−1)/(1−log _(10) (2))) = ((log _(10) (7)^2 +log _(10) (2)−1)/(1−log _(10) (2))) = ((2y+x−1)/(1−x)).

$$\:\begin{cases}{\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2}\right)=\mathrm{x}}\\{\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{7}\right)=\mathrm{y}}\end{cases}\Rightarrow\:\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{9}.\mathrm{8}\right)=? \\ $$$$\Leftrightarrow\:\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{9}.\mathrm{8}\right)=\:\frac{\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{98}\right)−\mathrm{1}}{\mathrm{1}−\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2}\right)} \\ $$$$\:=\:\frac{\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{7}\right)^{\mathrm{2}} +\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2}\right)−\mathrm{1}}{\mathrm{1}−\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2}\right)} \\ $$$$\:=\:\frac{\mathrm{2y}+\mathrm{x}−\mathrm{1}}{\mathrm{1}−\mathrm{x}}. \\ $$

Commented by mathdanisur last updated on 25/Jun/21

cool thanks Sir

$${cool}\:{thanks}\:{Sir} \\ $$

Answered by Rasheed.Sindhi last updated on 25/Jun/21

log 2=x log 7=y log_5 9.8=((log 9.8 )/(log 5))=((log(((98)/(10))))/(log(((10)/2)) ))=((log 98−log 10)/(log 10−log 2)) =((log 98−1)/(1−x)) =((log(2×7^2 ) −1)/(1−x))=((log 2+2log 7−1)/(1−x)) =((x+2y−1)/(1−x))

$$\mathrm{log}\:\mathrm{2}={x}\: \\ $$$$\mathrm{log}\:\mathrm{7}={y} \\ $$$$\mathrm{log}_{\mathrm{5}} \mathrm{9}.\mathrm{8}=\frac{\mathrm{log}\:\mathrm{9}.\mathrm{8}\:}{\mathrm{log}\:\mathrm{5}}=\frac{\mathrm{log}\left(\frac{\mathrm{98}}{\mathrm{10}}\right)}{\mathrm{log}\left(\frac{\mathrm{10}}{\mathrm{2}}\right)\:}=\frac{\mathrm{log}\:\mathrm{98}−\mathrm{log}\:\mathrm{10}}{\mathrm{log}\:\mathrm{10}−\mathrm{log}\:\mathrm{2}} \\ $$$$=\frac{\mathrm{log}\:\mathrm{98}−\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$=\frac{\mathrm{log}\left(\mathrm{2}×\mathrm{7}^{\mathrm{2}} \right)\:−\mathrm{1}}{\mathrm{1}−{x}}=\frac{\mathrm{log}\:\mathrm{2}+\mathrm{2log}\:\mathrm{7}−\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$=\frac{{x}+\mathrm{2}{y}−\mathrm{1}}{\mathrm{1}−{x}} \\ $$

Commented by mathdanisur last updated on 25/Jun/21

cool thank you Sir

$${cool}\:{thank}\:{you}\:{Sir} \\ $$

Answered by justtry last updated on 25/Jun/21

⇔((log(9,8))/(log5))=((log(((49×2)/(10))))/(log(((10)/2))))=((log7^2 +log2−log10)/(log10−log2)) =((2log7+log2−log10)/(log10−log2)) =((2y+x−1)/(1−x))

$$\Leftrightarrow\frac{{log}\left(\mathrm{9},\mathrm{8}\right)}{{log}\mathrm{5}}=\frac{{log}\left(\frac{\mathrm{49}×\mathrm{2}}{\mathrm{10}}\right)}{{log}\left(\frac{\mathrm{10}}{\mathrm{2}}\right)}=\frac{{log}\mathrm{7}^{\mathrm{2}} +{log}\mathrm{2}−{log}\mathrm{10}}{{log}\mathrm{10}−{log}\mathrm{2}} \\ $$$$=\frac{\mathrm{2}{log}\mathrm{7}+{log}\mathrm{2}−{log}\mathrm{10}}{{log}\mathrm{10}−{log}\mathrm{2}} \\ $$$$=\frac{\mathrm{2}{y}+{x}−\mathrm{1}}{\mathrm{1}−{x}} \\ $$

Commented by mathdanisur last updated on 25/Jun/21

Cool thanks Sir

$${Cool}\:{thanks}\:{Sir} \\ $$

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