Exponential Function: 4x+6x=9x

Let x be a real number that satisfy the equation

4x+6x=9x

What is the value of (23)x ?

Dividing both sides by 6x, you get
4x/6x+1=9x/6x
Simplifying the fractions you get
2x/3x+1=3x/2x
That can be written as
(2/3)x+1=(3/2)x
Now, if the number you want is (2/3)x=k, then (3/2)x=1/k ,
which means the equation can be written as
k+1=1/k
Multiplying both sides times k,
k2+k=1
Solving that quadratic equation gives you two real solutions for k.

I am not used anymore to calculate this type of equation, the result of relying too much in SHIFT + SOLVE of Casio. I did not say doing Shift + Solve is bad, it is actually highly recommended in my line of work, hehehe. Anyway, allow me to solve this, and I am actually surprised that solving for x is more complex than solving for (2/3)x. Here is my take based on the suggestion of KMST.
4x+6x=9x

4x6x+6x6x=9x6x

(46)x+1=(96)x

(23)x+1=(32)x

(23)x+1=1(23)x

(23)2x+(23)x=1

(23)2x+(23)x1=0
 

By Quadratic Equation
(23)x=1±124(1)(1)2(1)

(23)x=1±52
 

For
(23)x=1+52

log(23)x=log1+52

xlog(23)=log(51)log2

x(log2log3)=log(51)log2

x=log(51)log2log2log3   ←   a real number
 

For
(23)x=152

log(23)x=log152

xlog(23)=log(51)log2

x(log2log3)=log(51)log2

x=log(51)log2log2log3   ←   underfined

 

Hence,
(23)x=512   ←   this is my answer.

Please be gentle with me, hehehe. Although I hope I was able to consider everyhting, maybe I miss simething. As I've said, I am no longer used to this type of approach. A decimal number from calculator is more than acceptable for me.