Given-80-a-5-and-80-b-2-then-25-1-a-2b-1-a-4b-

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

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

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

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

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