How do you solve #5e ^ { 5x } = 1890#?

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Answer 1 · 2017-04-18T20:06:05

#x approx1.864#

The Equation is #5e^(5x)=1890#

Dividing by 5 on both sides we get :

#e^(5x)=378#

Take natural log on both sides:

# ln(e^(5x))=ln(378) #

( using property #ln(a^b)=bln(a)# )

#5xln(e)=ln(378) #

As #ln(e)=1#

#5x=ln (378) #

#5x=ln(2*3*3*3*7)# (factorized 378)

(using property # ln(a*b*c)=ln(a) + ln(b) + ln(c)# )

#5x=ln(2) +3ln(3) + ln(7)#

#x=1/5 (ln(2) +3ln(3) + ln(7))#

# x= 1/5 (.693 + 3(1.098) + 1.945) #

# x approx1.864 #