How do you solve $e^ { - 4x } + 8= 35$ ?

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Answer 1 · 2017-05-18T23:42:57

You have to use logarithms.

$e^(-4x) + 8 = 35$

$e^(-4x) = 27$

$ln 27 = - 4x$

$(ln 27) / (-4) = x$

$"*"ln$ is basically $log_e$

$x$ is approximately $-0.8240$ .

How do you solve e^ { - 4x } + 8= 35e^ { - 4x } + 8= 35?