How do you solve $7 = 5 e^{0.2 x}$ $7=5e^(0.2x)$ ?

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Answer 1 · 2015-12-13T18:04:35

$x = 5 ln (\frac{7}{5}) = ln (\frac{16807}{3125})$ $x=5ln(7/5)=ln(16807/3125)$

$[1] 7 = 5 e^{0.2 x}$ $[1]" "7=5e^(0.2x)$

Divide both sides by 5.

$[2]" "7/5=(cancel5e^(0.2x))/cancel5$

$[3]" "7/5=e^(0.2x)$

Change to log form.

$[4]" "hArrlog_(e)(7/5)=0.2x$

$[5]" "ln(7/5)=0.2x$

Divide both sides by 0.2.

$[6]" "ln(7/5)/0.2=(cancel0.2x)/cancel0.2$

$[7]" "x=ln(7/5)/0.2$

Simplify.

$[8]" "x=ln(7/5)/(1/5)$

$[9]" "color(blue)(x=5ln(7/5))$

This is optional, but you can continue by doing this.

Property of log: $nlog_ba=log_ba^n$

$[10]" "x=ln(7/5)^5$

$[11]" "color(blue)(x=ln(16807/3125))$

How do you solve 7=5e^(0.2x)7=5e0.2x7=5e^(0.2x)?