How do you solve #2 ^(4x)= 2.7#?

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Answer 1 · 2018-04-20T09:51:03

#color(blue)(x=ln(2.7)/(4ln(2))~~0.3582398518)#

By the laws of logarithms:

#log_a(b^c)=clog_a(b)#

#2^(4x)=2.7#

Taking natural logarithms of both sides:

#4xln(2)=ln(2.7)#

Divide by #4ln(2)#:

#x=ln(2.7)/(4ln(2))~~0.3582398518#