1. ## what is x

for what value of x is this possible
4^x + 6^x=9^x

2. Since this is posted under "Number Theory" is x assumed to be a positive integer? If so, there is no such x. $\displaystyle 4^x+ 6^x< 9^x$ for all x> 1. since $\displaystyle 4^1+ 6^2= 10> 9^1$, there is a root between 1 and 2.

3. Hello, jashansinghal!

For what value of $\displaystyle x$ is this possible? . $\displaystyle 4^x + 6^x \:=\:9^x$
We have: .$\displaystyle 9^x - 6^x - 4^x \:=\:0$

Divide by $\displaystyle 4^x\!:\quad \frac{9^x}{4^x} - \frac{6^x}{4^x} - \frac{4^x}{4^x} \:=\:0 \quad\Rightarrow\quad \left(\frac{9}{4}\right)^x - \left(\frac{6}{4}\right)^x - 1 \;=\;0$

. . $\displaystyle \left[\left(\frac{3}{2}\right)^2\right]^x - \left(\frac{3}{2}\right)^x - 1 \;=\;0 \quad\Rightarrow\quad \left[\left(\frac{3}{2}\right)^x\right]^2 - \left(\frac{3}{2}\right)^x - 1 \;=\;0$

. . . We have a quadratic in $\displaystyle \left(\frac{3}{2}\right)^x$

Quadratic Formula: .$\displaystyle \left(\frac{3}{2}\right)^x \;=\;\frac{1 \pm\sqrt{5}}{2}$

Since $\displaystyle \left(\frac{3}{2}\right)^x > 0$, the only root is: .$\displaystyle \left(\frac{3}{2}\right)^x \:=\:\frac{1+\sqrt{5}}{2} \:=\:\phi$, the Golden Mean.

So we have: .$\displaystyle \left(\frac{3}{2}\right)^x \:=\:\phi \quad\Rightarrow\quad (1.5)^x \:=\:\phi$

Take logs: .$\displaystyle \ln(1.5)^x \:=\:\ln(\phi) \quad\Rightarrow\quad x\ln(1.5) \:=\:\ln(\phi)$

Therefore: .$\displaystyle x \;=\;\frac{\ln(\phi)}{\ln(1.5)} \;=\;1.18681439\hdots$