How do you solve #5^(x-1) = 2^x#?

1 Answer
Aug 5, 2015

I found: #x=(ln5)/(ln5-ln2)#

Explanation:

I would take the natural log (#ln#) on both sides:
#ln5^(x-1)=ln2^x#
then use the fact that:
#lnx^a=alnx#
to get:
#(x-1)ln5=xln2#
#xln5-ln5-xln2=0#
#x(ln5-ln2)=ln5#
so that:
#x=(ln5)/(ln5-ln2)#
This is equal to #=1.7564# (if you can use the calculator).