How do you solve $e^(3r)+4=59$ ?

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Answer 1 · 2017-01-28T05:55:43

$r=ln55/3=1.336$

$e^(3r)+4=59$ can be rewritten as $e^(3r)=59-4=55$

As " $ln$ " is natural logarithm to the base $e$ ,

by taking logarithm of both sides to base $e$ ,

we get $" "3r=ln55$ and

$r=ln55/3=4.00733/3=1.33578$ or $1.336$

(using scientific calculator)

How do you solve e^(3r)+4=59e^(3r)+4=59?