Question

Solve for x :

$6^x+4^x=9^x$

Answer 1

$4^x+6^x=9^x$

On dividing both sides by $4^x$ we get

$1+{\left(\frac{6}{4}\right)}^x={\left(\frac{9}{4}\right)}^x$

$1+{\left(\frac{3}{2}\right)}^x={\left({\left(\frac{3}{2}\right)}^2\right)}^x$

${\left({\left(\frac{3}{2}\right)}^2\right)}^x-{\left(\frac{3}{2}\right)}^x-1=0$

${\left({\left(\frac{3}{2}\right)}^x\right)}^2-{\left(\frac{3}{2}\right)}^x-1=0$

Let us assume that ${\left(\frac{3}{2}\right)}^x=y$ the equation becomes,

$y^2-y-1=0...........................\left(i\right)$

Using quadratic formula i.e. $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing equation (i) with the general form of the quadratic equation we get $a=1,b=-1,c=-1$

$y=\frac{1\pm \sqrt{{\left(-1\right)}^2-4\times 1\times -1}}{2\times 1}$

$y=\frac{1\pm \sqrt{1+4}}{2}$

$y=\frac{1\pm \sqrt{5}}{2}$

${\left(\frac{3}{2}\right)}^x=\frac{1\pm \sqrt{5}}{2}$(taking log both sides we get)

$x\log \left(\frac{3}{2}\right)=\log \left(\frac{1\pm \sqrt{5}}{2}\right)$(as log of $2$ is not defined)

$x=\frac{\log \left({\displaystyle \frac{1\pm \sqrt{5}}{2}}\right)}{\log \left({\displaystyle \frac{3}{2}}\right)}$