How do you solve #5e^(2x)=3#?

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Answer 1 · 2018-03-06T19:29:47

#~~-0.2554128119#

The laws of logarithms state that:

1) #log_ab^c=clog_ab#

2) #log_aa=1#

#5e^2x=3#

Divide by #5#:

#e^(2x)=3/5#

Taking natural logarithms of both sides:

#2xlne=ln(3/5)color(white)(8888)# by 1

By 2:

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

Divide by 2:

#x=ln(3/5)/2~~-0.2554128119#