Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

Question Number 180014 by cortano1 last updated on 06/Nov/22

Given 80^a = 5 and 80^b = 2 then 25^((1−a−2b)/(1+a−4b)) =?

$$\:\:\mathrm{Given}\:\mathrm{80}^{{a}} \:=\:\mathrm{5}\:\mathrm{and}\:\mathrm{80}^{{b}} \:=\:\mathrm{2} \\ $$$$\:\mathrm{then}\:\mathrm{25}^{\frac{\mathrm{1}−{a}−\mathrm{2}{b}}{\mathrm{1}+{a}−\mathrm{4}{b}}} \:=?\: \\ $$

Answered by srikanth2684 last updated on 06/Nov/22

25^((ln_(80) 80−ln_(80) 5−2ln_(80) 2)/(ln_(80) 80+ln_(80) 5−4ln_(80) 2)) =25^((ln_(80) ((80)/(5×4)))/(ln_(80) ((80×5)/(16)))) =25^((ln_(80) 2^2 )/(ln_(80) 5^2 )) =(5^2 )^(ln_5^2 2^2 ) =4

$$\mathrm{25}^{\frac{\mathrm{ln}_{\mathrm{80}} \mathrm{80}−\mathrm{ln}_{\mathrm{80}} \mathrm{5}−\mathrm{2ln}_{\mathrm{80}} \mathrm{2}}{\mathrm{ln}_{\mathrm{80}} \mathrm{80}+\mathrm{ln}_{\mathrm{80}} \mathrm{5}−\mathrm{4ln}_{\mathrm{80}} \mathrm{2}}} \\ $$$$=\mathrm{25}^{\frac{\mathrm{ln}_{\mathrm{80}} \frac{\mathrm{80}}{\mathrm{5}×\mathrm{4}}}{\mathrm{ln}_{\mathrm{80}} \frac{\mathrm{80}×\mathrm{5}}{\mathrm{16}}}} =\mathrm{25}^{\frac{\mathrm{ln}_{\mathrm{80}} \mathrm{2}^{\mathrm{2}} }{\mathrm{ln}_{\mathrm{80}} \mathrm{5}^{\mathrm{2}} }} \\ $$$$=\left(\mathrm{5}^{\mathrm{2}} \right)^{{ln}_{\mathrm{5}^{\mathrm{2}} } \mathrm{2}^{\mathrm{2}} } =\mathrm{4} \\ $$

Commented by cortano1 last updated on 06/Nov/22

yes.

$$\mathrm{yes}.\: \\ $$

Answered by a.lgnaoui last updated on 06/Nov/22

80^a =5 a×log80=log5 a=((log5)/(log5+4log2)) 80^b =2 b×log80=log2 b=((log2)/(log5+4log2)) =25^((1−a−2b)/(1+a−4b)) =25^((1−((log5)/(log5+4log2))− ((2log2)/(log5+4log2)))/(1+((log5)/(log5+4log2))−((4log2)/(log5+4log2)))) =25^((log5+4log2−log5−2log2)/(log5+4log2+log5−4log2)) =25^((log2)/(log5)) ((log2)/(log5))×2log5=2log2 finaly: 25^((1−a−2b)/(1+a−4b)) =e^(2log2)

$$\mathrm{80}^{{a}} =\mathrm{5}\:\:\:\:\:{a}×\mathrm{log80}=\mathrm{log5}\:\:\:\:\:\:\:\:\mathrm{a}=\frac{\mathrm{log5}}{\mathrm{log5}+\mathrm{4log2}} \\ $$$$\mathrm{80}^{\mathrm{b}} =\mathrm{2}\:\:\:\:\mathrm{b}×\mathrm{log80}=\mathrm{log2}\:\:\:\:\mathrm{b}=\frac{\mathrm{log2}}{\mathrm{log5}+\mathrm{4log2}} \\ $$$$ \\ $$$$=\mathrm{25}^{\frac{\mathrm{1}−\mathrm{a}−\mathrm{2b}}{\mathrm{1}+\mathrm{a}−\mathrm{4b}}} =\mathrm{25}^{\frac{\mathrm{1}−\frac{\mathrm{log5}}{\mathrm{log5}+\mathrm{4log2}}−\:\frac{\mathrm{2log2}}{\mathrm{log5}+\mathrm{4log2}}}{\mathrm{1}+\frac{\mathrm{log5}}{\mathrm{log5}+\mathrm{4log2}}−\frac{\mathrm{4log2}}{\mathrm{log5}+\mathrm{4log2}}}} \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{25}^{\frac{\mathrm{log5}+\mathrm{4log2}−\mathrm{log5}−\mathrm{2log2}}{\mathrm{log5}+\mathrm{4log2}+\mathrm{log5}−\mathrm{4log2}}} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{25}^{\frac{\mathrm{log2}}{\mathrm{log5}}} \\ $$$$\:\:\:\frac{\mathrm{log2}}{\mathrm{log5}}×\mathrm{2log5}=\mathrm{2log2} \\ $$$$\:\:\:\mathrm{finaly}: \\ $$$$\:\:\:\:\:\:\:\mathrm{25}^{\frac{\mathrm{1}−\mathrm{a}−\mathrm{2b}}{\mathrm{1}+\mathrm{a}−\mathrm{4b}}} =\mathrm{e}^{\mathrm{2log2}} \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 06/Nov/22

e^(2log2) is synonyme of 25^((log2)/(log5)) =4

$${e}^{\mathrm{2log2}} \:\mathrm{is}\:\mathrm{synonyme}\:\mathrm{of}\:\:\:\mathrm{25}^{\frac{\mathrm{log2}}{\mathrm{log5}}} =\mathrm{4} \\ $$

Commented by a.lgnaoui last updated on 06/Nov/22

log2=log_e 2=((log2)/(loge))=((log2)/1)

$$\mathrm{log2}=\mathrm{log}_{\mathrm{e}} \mathrm{2}=\frac{\mathrm{log2}}{\mathrm{loge}}=\frac{\mathrm{log2}}{\mathrm{1}} \\ $$

Commented by a.lgnaoui last updated on 06/Nov/22

Resultat: 4

$${Resultat}:\:\mathrm{4} \\ $$

Commented by cortano1 last updated on 06/Nov/22

log 2= log _e (2)) ?

$$\left.\mathrm{log}\:\mathrm{2}=\:\mathrm{log}\:_{\mathrm{e}} \left(\mathrm{2}\right)\right)\:? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com