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Logarithm question is $\log_9(\sqrt[3]{(x^2-1)^2}) = \frac{1}{3}$

When I solved the question I rewrote the question as $\sqrt[3]{9} = \sqrt[3]{(x^2-1)^2}$, canceled the cube roots and then took the square root of both sides to get $x^2-1 = \pm3$ and then getting $x^2 = 4, x^2 = -2$, giving the solutions of $2, -2, \sqrt{2}i, -\sqrt{2}i,$ however after putting the question into online calculators like wolfram alpha, they only gave the solutions of $2, -2.$

chair
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2 Answers2

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Many calculators only know how to calculate real solutions. When I typed your formula into Wolfram Alpha (using cbrt to indicate cube root) it said:

assuming "cbrt" is the real-valued root

I selected the option to "Use the principal root instead" and it gave me all four answers as expected.

At first, I thought it had to do with $\log$ operating on real numbers; however, this is not correct because even if $x$ is complex, $x^2-1$ is $3$ when $x=2,-2$ and $-3$ when $x=i\sqrt2,-i\sqrt2$ so $(x^2-1)^2$ is always $9$ so the value under the cube root is always real anyway and so there is a real root to $\log$.

So, I think the issue is just with the calculator itself. I don't know how you typed it into Wolfram Alpha, but I got an option to use the principal root and it made the two complex solutions appear as well.

hyper-neutrino
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You have to remember that $\log_a(x)$ with $a>0\land a\neq1$ is a function of real number. This meams that letting $f(x)=\log_a(x)$: $$f: \mathbb{R}\to \mathbb{R}$$ Or, in other words, $\mathbb{R}$ is the domain and codomai of $\log_a(x)$. You can define $\log$ function in complex numbers, but it's another tool. So, WolpramAlpha is right, your equation: $$\log_9(\sqrt[3]{x^2-1})=\frac{1}{3}$$ has only two solutions: $x_1=2 \lor x_2=-2$.

Here a basic rule: when you are trying to solve this type of equation, always look carefully to the domain that is in this case: $$x\in (-\infty, -1) \cup(1, +\infty)$$

Matteo
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  • Even if $x$ is complex, in the case of $i\sqrt2$, $x^2=-2$ so $x^2-1=-3$ so $(x^2-1)^2=9$ and so $\log\left(\sqrt[3]{(x^2-1)^2}\right)$ should work because there is a real root of $\sqrt[3]{9}$ which works under $\log$, wouldn't it? – hyper-neutrino Nov 22 '21 at 21:31
  • @hyper-neutrino: we are working with real number (cube root or logarithm do not matter). All function are of the type: $\mathbb{R}\to\mathbb{R}$. – Matteo Nov 22 '21 at 21:33
  • If OP is getting complex numbers in their answer and are asking about it, clearly they aren't working with everything being real. Even with the real defn of logarithm and cube root it still doesn't explain why $x$ itself can't be complex. – hyper-neutrino Nov 22 '21 at 21:35
  • I understood that the domain of the log function needs to be real numbers but the x in the answers that is complex isn't the input of the log function but is just inside the input and you will get the cube root of 9 as the input as hyper-neutrino said – chair Nov 22 '21 at 21:47