Question

How to solve: 1" title="log_{3} (x+6) + log_{3}(x-6) = 4" alt="log_{3} (x+6) + log_{3}(x-6) = 4" align="absmiddle" class="latex-formula">

Answer 1

Collapse the logarithms into one:

$\log_3(x+6)+\log_3(x-6)=\log_3((x+6)(x-6))=4$

Then write both sides as powers of 3:

$3^{\log_3((x+6)(x-6))}=3^4$

$(x+6)(x-6)=81$

Simplify the left side and solve for $x$:

$x^2-36=81$

$x^2=117$

$x=\pm\sqrt{117}=\pm3\sqrt{13}$

But we're not done yet. Notice that both $x+6$ and $x-6$ are negative when $x=-3\sqrt{13}$. We can't take the logarithm of a negative number (the result isn't real-valued, anyway), so we throw this solution out, and we're left with just

$x=3\sqrt{13}$