[tex]\begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \underset{\stackrel{\uparrow }{\textit{let's use this one}}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\log_4(x+10)-\log_4(x-2)=\log_4(x)\implies \log_4\left( \cfrac{x+10}{x-2} \right)=\log_4(x) \\\\\\ \stackrel{\textit{exponentializing both sides}}{4^{\log_4\left( \frac{x+10}{x-2} \right)}=4^{\log_4(x)}}\implies \cfrac{x+10}{x-2}=x\implies x+10=x^2-2x \\\\\\ 10=x^2-3x\implies 0=x^2-3x-10 \\\\\\ 0=(x-5)(x+2)\implies x= \begin{cases} 5~~\checkmark\\ -2 \end{cases}[/tex]
notice, -2 is a valid value for the quadratic, however, the argument value for a logarithm can never 0 or less, it has to be always greater than 0, so for the logarithmic expression with (x-2), using x = -2 will give us a negative value, so -2 is no dice.
