How do you solve #6^x + 4^x = 9^x#?

Question

67207 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-17T12:07:22

#x=(ln((1+sqrt(5))/2))/(ln (3/2))#

Divide by #4^x# to form a quadratic in #(3/2)^x#.
Use #6^x/4^x=(6/4)^x=(3/2)^x and (9/4)^x=((3/2)^2)^x=((3/2)^x)^2#.
#((3/2)^x)^2-(3/2)^x-1=0#

So,# (3/2)^x=(1+-sqrt(1-4*1*(-1)))/2=(1+-sqrt(5))/2#

For the positive solution:

# (3/2)^x=(1+sqrt(5))/2#

Applying logarythms:

#xln (3/2)=ln((1+sqrt(5))/2)#

#x=(ln((1+sqrt(5))/2))/(ln (3/2))=1.18681439....#