How do you solve $12^{y - 2} = 20$ $12^(y-2) =20$ ?

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Answer 1 · 2016-08-23T22:25:31

$y = \frac{ln 2880}{ln 12}$ $y = ln2880/ln12$

Take the natural logarithm of both sides.

$ln (12^{y - 2}) = ln 20$ $ln(12^(y - 2)) = ln20$

$(y - 2) ln 12 = ln 20$ $(y - 2)ln12 = ln20$

$y ln 12 - 2 ln 12 = ln 20$ $yln12 - 2ln12 = ln20$

$y ln 12 = ln 20 + ln 144$ $yln12 = ln20 + ln144$

$y = \frac{ln 2880}{ln 12}$ $y = ln2880/ln12$

Hopefully this helps!

How do you solve 12y−2=2012^(y-2) =20?