Solve #2^x+3^x = 4^x# ?

1 Answer
Oct 17, 2017

See below.

Explanation:

We have

#2^1+3^1 = 5 > 4^1 = 4# and
#2^2+3^2 = 13 < 4^2 = 16#

Then by continuity there exists at least a #x = x_0 in [1,2]#
such that #2^(x_0)+3^(x_0) = 4^(x_0)#

Using an iterative process we can determine

#x_0 approx 1.507126591638653#