How do you solve #4(2^x) = 5^x#?

1 Answer
Gió
Nov 18, 2015

I found: #x=1.5129#

Explanation:

Start by writing it as:
#2^2*2^x=5^x#
#2^(x+2)=5^x#
take the #ln# of both sides:
#ln(2^(x+2))=ln5^x#
use the property of logs:
#logx^a=alogx#
to get:
#(x+2)ln2=xln5#
#xln2+2ln2-xln5=0#
isolate #x#:
#x(ln2-ln5)=-2ln2#
#x=-(2ln2)/(ln2-ln5)=-1.3863/-0.9163=1.5129#

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