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Question Number 218279 by Tawa11 last updated on 04/Apr/25

Evaluate: (4^(log_(5/4) 4) /5^(log_(5/4) 5) ) Show workings please.

$$\mathrm{Evaluate}: \\ $$$$\:\:\:\:\:\frac{\mathrm{4}^{\mathrm{log}_{\frac{\mathrm{5}}{\mathrm{4}}} \mathrm{4}} }{\mathrm{5}^{\mathrm{log}_{\frac{\mathrm{5}}{\mathrm{4}}} \mathrm{5}} } \\ $$$$\mathrm{Show}\:\mathrm{workings}\:\mathrm{please}. \\ $$

Answered by Frix last updated on 05/Apr/25

a^(log_b c) =a^((ln c)/(ln b)) =e^((ln a ln c)/(ln b)) ln (m/n) =ln m −ln n ⇒ 4^(log_(5/4) 4) =4^((ln 4)/(ln 5 −ln 4)) =e^((ln^2 4)/(ln 5 −ln 4)) 5^(log_(5/4) 5) =5^((ln 5)/(ln 5 −ln 4)) =e^((ln^2 5)/(ln 5 −ln 4)) =^((∗)) =e^(((ln^2 4)/(ln 5 −ln 4))+ln 5 +ln 4) =e^((ln^2 4)/(ln 5 −ln 4)) e^(ln 5) e^(ln 4) = =20e^((ln^2 4)/(ln 5 −ln 4)) ⇒ answer is (1/(20)) (∗) (a^2 /(a−b))=((a^2 −b^2 +b^2 )/(a−b))=(((a−b)(a+b)+b^2 )/(a−b))= =a+b+(b^2 /(a−b))

$${a}^{\mathrm{log}_{{b}} \:{c}} ={a}^{\frac{\mathrm{ln}\:{c}}{\mathrm{ln}\:{b}}} =\mathrm{e}^{\frac{\mathrm{ln}\:{a}\:\mathrm{ln}\:{c}}{\mathrm{ln}\:{b}}} \\ $$$$\mathrm{ln}\:\frac{{m}}{{n}}\:=\mathrm{ln}\:{m}\:−\mathrm{ln}\:{n} \\ $$$$\Rightarrow \\ $$$$\mathrm{4}^{\mathrm{log}_{\frac{\mathrm{5}}{\mathrm{4}}} \:\mathrm{4}} =\mathrm{4}^{\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{ln}\:\mathrm{5}\:−\mathrm{ln}\:\mathrm{4}}} =\mathrm{e}^{\frac{\mathrm{ln}^{\mathrm{2}} \:\mathrm{4}}{\mathrm{ln}\:\mathrm{5}\:−\mathrm{ln}\:\mathrm{4}}} \\ $$$$ \\ $$$$\mathrm{5}^{\mathrm{log}_{\frac{\mathrm{5}}{\mathrm{4}}} \:\mathrm{5}} =\mathrm{5}^{\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{ln}\:\mathrm{5}\:−\mathrm{ln}\:\mathrm{4}}} =\mathrm{e}^{\frac{\mathrm{ln}^{\mathrm{2}} \:\mathrm{5}}{\mathrm{ln}\:\mathrm{5}\:−\mathrm{ln}\:\mathrm{4}}} \:\overset{\left(\ast\right)} {=} \\ $$$$\:\:\:\:\:=\mathrm{e}^{\frac{\mathrm{ln}^{\mathrm{2}} \:\mathrm{4}}{\mathrm{ln}\:\mathrm{5}\:−\mathrm{ln}\:\mathrm{4}}+\mathrm{ln}\:\mathrm{5}\:+\mathrm{ln}\:\mathrm{4}} =\mathrm{e}^{\frac{\mathrm{ln}^{\mathrm{2}} \:\mathrm{4}}{\mathrm{ln}\:\mathrm{5}\:−\mathrm{ln}\:\mathrm{4}}} \mathrm{e}^{\mathrm{ln}\:\mathrm{5}} \mathrm{e}^{\mathrm{ln}\:\mathrm{4}} = \\ $$$$\:\:\:\:\:=\mathrm{20e}^{\frac{\mathrm{ln}^{\mathrm{2}} \:\mathrm{4}}{\mathrm{ln}\:\mathrm{5}\:−\mathrm{ln}\:\mathrm{4}}} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{20}} \\ $$$$\left(\ast\right)\:\frac{{a}^{\mathrm{2}} }{{a}−{b}}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}−{b}}=\frac{\left({a}−{b}\right)\left({a}+{b}\right)+{b}^{\mathrm{2}} }{{a}−{b}}= \\ $$$$\:\:\:\:\:\:\:={a}+{b}+\frac{{b}^{\mathrm{2}} }{{a}−{b}} \\ $$

Commented by Tawa11 last updated on 05/Apr/25

Thanks sir. I really appreciate.

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Commented by Frix last updated on 05/Apr/25

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Answered by Frix last updated on 05/Apr/25

Generally (a^(ln_(b/a) a) /b^(ln_(b/a) b) )=(1/(ab))

$$\mathrm{Generally} \\ $$$$\frac{{a}^{\mathrm{ln}_{\frac{{b}}{{a}}} \:{a}} }{{b}^{\mathrm{ln}_{\frac{{b}}{{a}}} \:{b}} }=\frac{\mathrm{1}}{{ab}} \\ $$

Commented by Tawa11 last updated on 05/Apr/25

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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