How do you solve $2.1^(t-5)=9.32$ ?

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Answer 1 · 2016-12-10T12:56:27

I have taken you to a point where you can complete it

Using the principle that $log(a^b) -> blog(a)$

Take logs of both sides

$(t-5)log(2.1)=log(9.23)$

$t-5=(log(9.23))/(log(2.1))$

$t=(log(9.23))/(log(2.1))+5$

How do you solve 2.1^(t-5)=9.322.1^(t-5)=9.32?