Logarithmic equation 3^(x+1)+9^(x-1)-61236=0 produces unclear solution

Question

I have a basic logarithmic equation

3^(x+1) +9^(x-1) -61236 =0

which I simplified like

(x+1)*ln(3) + (2x-2)*ln(3) = ln(61236)

and furthermore

3x-1 = ln(61233)

If you solve for $x$ you get approximately $4.00748.$ If you enter the original equation into WolframAlpha the integer solution is $6.$ The WolframAlpha Link is here.

Where is my mistake? Thank you in advance!

Answer 1

You make a mistake when you take logarithms.

$\log a + \log b = \log ab$ and not $\log(a+b)=\log a+\log b$

Answer 2

Let $z=3^x$. Then your equation can be written as $$ 61236=3^{x+1}+\frac{1}{9}9^x=3z+\frac{1}{9}z^2\iff z=-756\quad\text{or}\quad z=729. $$ Is $z=-756$ admissible? What does $729=z=3^x$ imply about $x$?