4^x+6^x =9^x, find x

4^x＋6^x＝9^x

As 6^x＞0 for real x, we can divide the equation by 6^x and get

(4/6)^x＋1＝(9/6)^x
(2/3)^x＋1＝(3/2)^x

Let y＝(2/3)^x, we get

y＋1＝1/y
y^2＋y＝1
y^2＋y－1＝0

y＝(±√5－1)/2

As y＝(2/3)^x＞0 for real x, we get

y＝(√5－1)/2


As y＝(2/3)^x, we get

ln y ＝x ln (2/3)
x ＝ln y / ln (2/3)＝ ln [(√5－1)/2] / ln (2/3)＝1.1868

If you plot
4^x+6^x-9^x against x, you will notice a root at x=1.1865 (approx.). Zoomed in image at:
http://i263.photobucket.com/albums/ii157/mathmate/4x_6x-1.png

Using this information, you can refine the solution by rewriting the equation as
xlog9=log(4^x+6^x)
x=log(4^x+6^x)/log9
Starting with x=1.187 on the right hand side, and repeatedly substituting the new value of x on the RHS, we get successively:
x1=1.186952665878328
x2=1.186917402828997
x3=1.186891132532082
Using Shank's transformation on x1,x2 and x3, we obtain
x=(x1*x3-x2^2)/(x1+x3-2*x2)=1.186814389796409
accurate to 8 places of decimal.
The accurate solution is:
1.186814390280981