How do you solve #3.5^(x+2) = 1.75^(x+3)#?

1 Answer
Oct 5, 2015

#x = (ln(7/16))/ln(2) approx -1.19265#

Explanation:

Take the natural logarithm (or any logarithm) of both sides and use properties of logarithms to help you solve for #x#.

#ln(3.5^{x+2})=ln(1.75^{x+3})\rightarrow (x+2)ln(3.5)=(x+3)ln(1.75)#

#\rightarrow x(ln(1.75)-ln(3.5))=2ln(3.5)-3ln(1.75)#

#\rightarrow x=(ln(3.5^2)-ln(1.75^3))/(ln(1.75/3.5))=ln(16/7)/ln(1/2)#

#=ln(7/16)/ln(2) approx -1.19265#