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

Question

2004 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-18T23:42:57

You have to use logarithms.

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

$e^{- 4 x} = 27$ $e^(-4x) = 27$

$ln 27 = - 4 x$ $ln 27 = - 4x$

$\frac{ln 27}{- 4} = x$ $(ln 27) / (-4) = x$

$* ln$ $"*"ln$ is basically ${log}_{e}$ $log_e$

$x$ $x$ is approximately $- 0.8240$ $-0.8240$ .

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