How do you solve $ln (x - 7) = 2$ $ln(x-7)=2$ ?

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Answer 1 · 2016-10-30T05:15:58

$e^{2} + 7$ $e^2+7$

here is how you do it

$ln (x - 7) = 2$ $ln(x-7)=2$
$or, e^{ln (x - 7)} = e^{2}$ $or, e^(ln(x-7))=e^2$
$or, x - 7 = e^{2}$ $or, x-7=e^2$
$or, x = e^{2} + 7$ $or,x=e^2+7$

How do you solve ln(x-7)=2ln(x−7)=2ln(x-7)=2?