Logs going through a quadratic isn't a valid solution path?

Question

Consider $$\log_4(x+5)+\log_4(x+11)=2$$

Raising $4$ to each term across the board and treating the addition as a multiplication you get a simple quadratic, whose roots are $x=\{-3,-13\}$. Easy Peasy. But wait. A log function isn't a parabola; there's obviously only one solution! Wolfram Alpha, goes via a completely different path (which I leave the reader to expose) and ends up with only the $-3$ solution -- so I'm at least not completely crazy...or only half so, anyway! :-)

One way of thinking about why $-13$ isn't a legal solution is that if $x=-13$ then $\log_4(x+5)=\log_4(-8)$ which is imaginary (and ditto BTW for the second term). Okay, but I'm not sure why that's not a valid solution; roots can be imaginary. But what's wrong with the way I solved this?

What algebraic step is illegal in my solution based on anti-logging (exponentiation), and then solving the resulting quadratic? It's not like I'm dividing by zero. (Or if I am, I don't see where.)

Answer 1

Certainly you can get an irrational result from taking a logarithm. In problems that aren't carefully set up to have rational results, you very often (usually?) get an irrational answer.

But if you take the log of a negative number, the result isn't irrational; it is undefined. That is why $-13$ is not a solution.

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Addendum. The response above is the story for real analysis. When you ask Wolfram Alpha to solve $\log_4(x+5)+\log_4(x+11)=2,$ it probably assumes you mean real analysis.

If you are working in complex analysis, then the natural logarithm is defined for all numbers except zero. The complex logarithms of negative real numbers have only complex values, not rational real or even irrational real. On the other hand, the complex logarithm is multivalued, and mathematicians rarely if ever talk about complex logarithms other than the natural logarithm.

If you define the principal value of $\log_4$ in complex analysis by $$ \log_4(z) = \frac{\ln z}{\ln 4}, $$ taking the principal values of both $\ln z$ and $\ln 4,$ then $$ \log_4(-8) + \log_4(-2) = 2 + i\frac{\pi}{\ln 2} \neq 2. $$ The discrepancy is because $\ln(xy) = \ln x + \ln y$ is not true for the principal value of the natural logarithm in complex analysis.

Answer 2

The issue comes into play when you use the rule $a^{\log_ax}=x$. This equation is indeed true, but it only holds on an appropriately restricted domain: we must have $x$ positive. For negative or zero $x$, the LHS is simply undefined. To illustrate, the function $2^{\log_2 x}+1$ is exactly equal to $x+1$ but, somehow, doesn't have a root at $x=-1$, nor any roots at all, for that matter. Why? Because the function is only well defined at positive $x$.

So, when you use this rule to derive $4^{\log_{4}\left(x+5\right)}\cdot4^{\log_{4}\left(x+11\right)}=\left(x+5\right)\cdot\left(x+11\right)$, you do get a product of linear terms, and hence a quadratic, but only an incomplete section of the ascending arm, with $x>5$. The descending arm and the second root are not in the domain of the function.