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

1 Answer
Jul 5, 2015

Answer:

I found: #x=(2ln(5))/(ln(5)-ln(2))#

Explanation:

I would apply the natural logarithm, #ln#, to both sides:
#ln(2^x)=ln(5^(x-2))#
apply one of the rules of logs:
#xln(2)=(x-2)ln(5)#
#xln(2)=xln(5)-2ln(5)# rearranging:
#x[ln(5)-ln(2)]=2ln(5)#
and finally:
#x=(2ln(5))/(ln(5)-ln(2))#

Using a pocket calculator you can get #x=3.513#