How do you solve $e^(3-5x)=16$ ?

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Answer 1 · 2018-04-20T08:37:31

$color(blue)(x=(ln(16)-3)/-5~~0.0454822555)$

By the laws of logarithms:

$log_a(b^c)=clog_a(b) \ \ \ \ \ [1]$

$log_a(a)=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [2]$

$e^(3-5x)=16$

Taking natural logarithms of both sides:

Using $[1]$

$(3-5x)ln(e)=ln(16)$

By $[2]$

$3-5x=ln(16)$

Subtract $3$ :

$-5x=ln(16)-3$

Divide by $-5$ :

$color(blue)(x=(ln(16)-3)/-5~~0.0454822555)$

How do you solve e^(3-5x)=16e^(3-5x)=16?