How do you solve $5.07^(t-1) = 100$ ?

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Answer 1 · 2016-04-14T17:17:17

$t = 1 + 2/log 5.07$ =3.837, nearly.

$log a^m=m loga and log 10^n=n$

Equating logarithms,

$(t-1) log 5.07=log 100=log 10^2=2$
So, $t=1+2/log 5.07$ .

How do you solve 5.07^(t-1) = 1005.07^(t-1) = 100?