Question

A medical battery loses 5% of its power each week. How long will it take until only 20% of
its power remains.

Answer if you only know

Answer 1

$\star$ $20=100(0.95)^t$ 

$\star$ $0.2=0.95^t$ 

$\star$ $Log \ 0.2 = t \ log.95$ 

$\star$ $\dfrac{log0.2}{log.95} = t$ 

$\star$ $31.4 \ weeks = t$ 

$\star$ $Hence \ solved$