Solve $\log_4(x^2 - x - 4) = 2$. Identify any extraneous roots.

This should be pretty straightforward, but I'm quite confused. I went through this problem without much difficulty initially, equated the arguments on each side, then factored and solved like any other quadratic. Simple. answers were $$x = -2$$ or $$x = 3$$ and when I checked those answers I found that $$-2$$ was an extraneous root and so I rejected it as a solution as it would cause me to take the log of a negative number. Then I went to check $$3$$ and that was in the domain no problem. But when I subbed $$3$$ back into the original equation, LS did not equal RS. I tried to figure out where I went wrong for a while then I tried expressing the constant on the right side as a log with base $$4$$, and the result was $$x = 1$$. Seemed to make more sense but once again, when subbed into the original equation, it also turned out to cause me to take the log of a negative number. So I'm really stuck here on where in the world I'm going wrong in finding the solution. I've been looking and looking and I can't find any mistakes. Could use some help please and thanks. Would be greatly appreciated.

• It looks like you've assumed $x$ needs to be positive, when what actually needs to be positive (indeed, $16$) is the logarithm's argument, which is $x^2-x-4$.
– J.G.
Apr 1, 2021 at 20:05

The first time, you did the following:

$$\log_4(x^2 - x - 4) = 2$$ $$x^2 - x - 4 = 2$$ $$x^2 - x - 6 = 0$$

That gets you $$x = 3$$ or $$x = -2$$.

The problem is, you're not accounting for the $$\log(\cdot)$$. You equated the input on the left with the value on the right, but both sides of an equation should be thought of as outputs, so you can't just drop the $$\log(\cdot)$$.

I'm not sure what you did the second time, but I think you might have incorrectly substituted $$\log_4(4)$$ in for the $$2$$ on the RHS, which is incorrect, $$\log_4(4) = 1$$.

You can deal with the $$\log(\cdot)$$ by raising both sides into an exponent with base $$4$$, then $$4^{\cdot}$$ and $$\log_4(\cdot)$$ cancel out:

$$\log_4(x^2 - x - 4) = 2$$ $$4^{\log_4(x^2 - x - 4)} = 4^2$$ $$x^2 - x - 4 = 16$$

Then, solve the same way you did before:

$$x^2 - x - 20 = 0$$ $$(x-5)(x+4) = 0$$ $$x=5, \ x=-4$$ $$\log_4(5^2 - 5 - 4) = \log_4(16) = 2 \ \checkmark$$ $$\log_4((-4)^2 - (-4) - 4) = \log_4(16) = 2 \ \checkmark$$

You've miscalculated somewhere (indeed, @AmaanM has noted you kept the RHS as $$2$$ instead of $$4^2$$ when removing the $$\log_4$$). The original equation is equivalent to $$x^2-x-4=4^2=16$$, i.e. $$x\in\{-4,\,5\}$$. Both roots count.