How do you solve #4^( x + 2) = 20#?

1 Answer
Gió
May 28, 2016

I found: #x=(ln(20))/(ln(4))-2=0.16096#

Explanation:

I would first take the natural log of both sides:
#ln(4)^(x+2)=ln(20)#
then use a property of logs to write:
#(x+2)ln(4)=ln(20)#
rearrange:
#x+2=(ln(20))/(ln(4))#
and:
#x=(ln(20))/(ln(4))-2=0.16096#

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