How do you solve #7=5e^(0.2x)#?

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Answer 1 · 2015-12-13T18:04:35

#x=5ln(7/5)=ln(16807/3125)#

#[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))#