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

Given 80^{a} = 5 and 80^{b} = 2

then 25^{\frac{1 - a - 2 b}{1 + a - 4 b}} = ?

Answered by srikanth2684 last updated on 06/Nov/22

25^{\frac{\ln_{80} 80 - \ln_{80} 5 - {2 \ln}_{80} 2}{\ln_{80} 80 + \ln_{80} 5 - {4 \ln}_{80} 2}}

= 25^{\frac{\ln_{80} \frac{80}{5 \times 4}}{\ln_{80} \frac{80 \times 5}{16}}} = 25^{\frac{\ln_{80} 2^{2}}{\ln_{80} 5^{2}}}

= {(5^{2})}^{{l n}_{5^{2}} 2^{2}} = 4

Commented by cortano1 last updated on 06/Nov/22

yes .

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

80^{a} = 5 a \times \log 80 = \log 5 a = \frac{\log 5}{\log 5 + 4 \log 2}

80^{b} = 2 b \times \log 80 = \log 2 b = \frac{\log 2}{\log 5 + 4 \log 2}

= 25^{\frac{1 - a - 2 b}{1 + a - 4 b}} = 25^{\frac{1 - \frac{\log 5}{\log 5 + 4 \log 2} - \frac{2 \log 2}{\log 5 + 4 \log 2}}{1 + \frac{\log 5}{\log 5 + 4 \log 2} - \frac{4 \log 2}{\log 5 + 4 \log 2}}}

= 25^{\frac{\log 5 + 4 \log 2 - \log 5 - 2 \log 2}{\log 5 + 4 \log 2 + \log 5 - 4 \log 2}}

= 25^{\frac{\log 2}{\log 5}}

\frac{\log 2}{\log 5} \times 2 \log 5 = 2 \log 2

finaly :

25^{\frac{1 - a - 2 b}{1 + a - 4 b}} = e^{2 \log 2}

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

e^{2 \log 2} is synonyme of 25^{\frac{\log 2}{\log 5}} = 4

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

\log 2 = \log_{e} 2 = \frac{\log 2}{loge} = \frac{\log 2}{1}

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

R e s u l t a t : 4

Commented by cortano1 last updated on 06/Nov/22

\log 2 = \log_{e} (2)) ?