Why do some logarithmic equations have two solutions

logarithms

I was trying to find solutions for a high school math problem, but there was one thing I didn't quite understand.

There is a logarithmic identity that says that

$ln\:x^2=2\cdot ln \:x$

However, when solving an equation, the two different forms give different solutions
When graphing in Geogebra, or trying to solve with wolframalpha, $ 2 \cdot ln\:x=1$ has only a positive answer to the equation ($\sqrt e$)

But when using $ln\:x^2=1$, I also get the solutions for negative x ($\:\sqrt e, -\sqrt e)$.

If these forms are exactly the same, why do they give different solutions?

Best Answer

The correct identity is

$$\log x^2=2\log |x|.$$

Related Solutions

Imaginary Numbers – Exploring x^c = -x

Robert's answer tells you correctly how to find, given an $x$, a complex $c$ such that $x^c=-x$.

However, it is worth pointing out that you cannot find a single $c$ such $x^c=-x$ will hold for every $x$. There's no existing number that achieves this, and if you try to extend the number system by simply postulating that $c$ exists (which I think was the point of the question), you run into this: $$ 0 = \frac08 = \frac{-4+4}{8} = \frac{4^c+4}8 = \frac{2^c2^c+4}8 = \frac{(-2)(-2)+4}8 = \frac{4+4}8 = 1 $$ So if you want your $c$ to exist, but don't want to collapse everything into $0=1$, then you have to discard the exponentiation rule $(ab)^c=a^cb^c$. And without that rule, the result isn't really worthy of being called exponentiation in the first place.

I think you have been fooled by a common introduction to complex numbers, which goes like this: "There is no real number whose square is $-1$, but assume we had such a number anyway and call it $i$. Then we can calculate with the assumption $i^2=-1$, and, ta-dah! we have invented a new kind of numbers". This presentation, however, sweeps an important point under the carpet, namely that we had better be sure that assuming $i^2=-1$ doesn't allow us to prove falsehoods such as $0=1$ about the real numbers that we already know and love -- because if it does that means that we haven't invented a new kind of numbers but simply deceived ourselves.

For $i^2=-1$ it turns out that it's not too difficult to prove that it doesn't lead to nonsense. But even when this proof is shown, something is still swept under the carpet here, namely that if we assume $i^2=-1$ and the familiar rule that every number is either 0 or positive or negative (and the usual rules for positive and negative numbers), then we can still derive nonsense. So what's really true is that we can get a $\sqrt{-1}$ if and only if we're willing to drop what we know about positive and negative numbers (or "less than"/"greater than") when we're working with complex numbers.

Extending the concept of number is almost always connected to a loss of some rules that used to hold before the extension. How far we can push the extension depends on how many rules we're willing to let go of. So you can have your $c$, but the cost you must pay is the rules for how exponentiation works. And if you pay that cost, having $c$ becomes pretty useless anyway, so the whole expedition turns out to be pointless.

(But of course we couldn't have known that before we went and checked. The idea was good enough; it just didn't work. Perhaps your next one will.)

Inequality – Solving Logarithmic Inequality in Two Hours

We want to solve $$ \ x \log_{\log_{f(x)} (g(x)) } \left( h(x)\right) \geq 0$$

where $$f(x)=|x^2 - 3 | - 2,\quad g(x)=x^2 - 3|x| + 2,\quad h(x)= \dfrac{x^3 - |3x+2|}{x^3 - |3x-2|}$$

First of all, $x=0$ is not a solution since $f(0)=1$.

Comparing $x$ with $0$, $f(x)$ with $1$, $g(x)$ with $f(x)$, $h(x)$ with $1$, we have eight cases to consider :

Case 1 : $x\lt 0$ and $0\lt f(x)\lt 1$ and $f(x)\lt g(x)\lt 1$ and $h(x)\ge 1$

Case 2 : $x\lt 0$ and $f(x)\gt 1$ and $1\lt g(x)\lt f(x)$ and $h(x)\ge 1$

Case 3 : $x\lt 0$ and $0\lt f(x)\lt 1$ and $0\lt g(x)\lt f(x)$ and $0\lt h(x)\le 1$

Case 4 : $x\lt 0$ and $f(x)\gt 1$ and $g(x)\gt f(x)$ and $0\lt h(x)\le 1$

Case 5 : $x\gt 0$ and $0\lt f(x)\lt 1$ and $f(x)\lt g(x)\lt 1$ and $0\lt h(x)\le 1$

Case 6 : $x\gt 0$ and $f(x)\gt 1$ and $1\lt g(x)\lt f(x)$ and $0\lt h(x)\le 1$

Case 7 : $x\gt 0$ and $0\lt f(x)\lt 1$ and $0\lt g(x)\lt f(x)$ and $h(x)\ge 1$

Case 8 : $x\gt 0$ and $f(x)\gt 1$ and $g(x)\gt f(x)$ and $h(x)\ge 1$

Using the following lemmas :

Lemma 1 : If $x\lt 0$, then $h(x)\lt 1$.

Lemma 2 : There are no $x$ such that $f(x)\gt 1$ and $g(x)\gt f(x)$.

(The proofs for the lemmas are written at the end of the answer.)

we see that Case 1, Case 2, Case 4, Case 8 don't happen.

Now, we have four cases to consider :

Case 3 : $x\lt 0$ and $0\lt g(x)\lt f(x)\lt 1$ and $0\lt h(x)\le 1$

Case 5 : $x\gt 0$ and $0\lt f(x)\lt g(x)\lt 1$ and $0\lt h(x)\le 1$

Case 6 : $x\gt 0$ and $1\lt g(x)\lt f(x)$ and $0\lt h(x)\le 1$

Case 7 : $x\gt 0$ and $0\lt g(x)\lt f(x)\lt 1$ and $h(x)\ge 1$

Case 3 : $x\lt 0$ and $0\lt g(x)\lt f(x)\lt 1$ and $0\lt h(x)\le 1$

Since $x\lt 0$, we have $g(x)=x^2-3(-x)+2=x^2+3x+2$.

Case 3-1 : For $x\in(-\infty,-\sqrt 3)$, we have $$f(x)=(x^2-3)-2=x^2-5,\quad h(x)=\frac{x^3-(-(3x+2))}{x^3-(-(3x-2))}=\frac{x^3+3x+2}{x^3+3x-2}$$ $$0\lt g(x)\lt f(x)\lt 1\iff 0\lt x^2+3x+2\lt x^2-5\lt 1\iff x\in\left(-\sqrt 6,-\frac 73\right)$$ Since $x^3+3x-2\lt 0$ for $x\lt 0$, $$0\lt h(x)\le 1\iff 0\lt \frac{x^3+3x+2}{x^3+3x-2}\le 1\iff x^3+3x-2\le x^3+3x+2\lt 0$$ $$\iff x^3+3x+2\lt 0\iff x\lt \alpha$$ where $\left(-\frac 73\lt\right)\alpha$ is the only root of $i(x)=x^3+3x+2$ which is increasing with $i(-7/3)\lt 0$.

So, in this case, we have $$x\in\left(-\sqrt 6,-\frac 73\right)\tag1$$

Case 3-2 : For $x\in\left[-\sqrt 3,-\frac 23\right)$, we have $$f(x)=-(x^2-3)-2=-x^2+1,\quad h(x)=\frac{x^3-(-(3x+2))}{x^3-(-(3x-2))}=\frac{x^3+3x+2}{x^3+3x-2}$$

$$0\lt g(x)\lt f(x)\lt 1\iff 0\lt x^2+3x+2\lt -x^2+1\lt 1\iff x\in\left(-1,-\frac 12\right)$$

$$0\lt h(x)\le 1\iff x\lt \alpha$$where $\left(-\frac 23\lt\right)\alpha$ is the only root of $i(x)=x^3+3x+2$ which is increasing with $i(-2/3)\lt 0$.

So, in this case, we have $$x\in\left(-1,-\frac 23\right)\tag2$$

Case 3-3 : For $x\in\left[-\frac 23,0\right)$, we have $$f(x)=-(x^2-3)-2=-x^2+1,\quad h(x)=\frac{x^3-(3x+2)}{x^3-(-(3x-2))}=\frac{x^3-3x-2}{x^3+3x-2}$$

$$0\lt g(x)\lt f(x)\lt 1\iff x\in\left(-1,-\frac 12\right)$$ $$0\lt h(x)\le 1\iff 0\lt \frac{x^3-3x-2}{x^3+3x-2}\le 1\iff x^3+3x-2\le x^3-3x-2\lt 0$$ $$\iff x^3-3x-2=(x-2)(x+1)^2\lt 0\iff x\in(-\infty,-1)\cup (-1,2)$$ So, in this case, we have $$x\in \left[-\frac 23,-\frac 12\right)\tag3$$

Therefore, in Case 3, we have $$(1)\cup (2)\cup (3)\iff x\in\left(-\sqrt 6,-\frac 73\right)\cup \left(-1,-\frac 12\right)\tag4$$

Case 5 : $x\gt 0$ and $0\lt f(x)\lt g(x)\lt 1$ and $0\lt h(x)\le 1$

Since $x\gt 0 $, we have $g(x)=x^2-3x+2$.

Case 5-1 : For $x\in\left(0,\frac 23\right)$, we have $$f(x)=-(x^2-3)-2=-x^2+1,\quad h(x)=\frac{x^3-(3x+2)}{x^3-(-(3x-2))}=\frac{x^3-3x-2}{x^3+3x-2}$$ $$0\lt f(x)\lt g(x)\lt 1\iff 0\lt -x^2+1\lt x^2-3x+2\lt 1\iff x\in\left(\frac{3-\sqrt 5}{2},\frac 12\right)$$ Since $x^3+3x-2\lt 0$ for $x\lt \frac 12$ where $j(x)=x^3+3x-2$ is increasing with $j(1/2)\lt 0$, $$0\lt h(x)\le 1\iff 0\lt \frac{x^3-3x-2}{x^3+3x-2}\le 1\iff x^3+3x-2\le x^3-3x-2\lt 0$$ which does not hold for $x\gt 0$.

Case 5-2 : For $x\in\left[\frac 23,\sqrt 3\right)$, we have $f(x)=-x^2+1$ and $$0\lt f(x)\lt g(x)\lt 1\iff x\in\left(\frac{3-\sqrt 5}{2},\frac 12\right)$$ There are no $x$ such that $x\in\left[\frac 23,\sqrt 3\right)\cap \left(\frac{3-\sqrt 5}{2},\frac 12\right)$.

Case 5-3 : For $x\in [\sqrt 3,\infty)$, we have $$f(x)=(x^2-3)-2=x^2-5,\quad h(x)=\frac{x^3-(3x+2)}{x^3-(3x-2)}=\frac{x^3-3x-2}{x^3-3x+2}$$ $$0\lt f(x)\lt g(x)\lt 1\iff 0\lt x^2-5\lt x^2-3x+2\lt 1\iff x\in\left(\sqrt 5,\frac 73\right)$$ Since $x^3-3x+2=(x+2)(x-1)^2\gt 0$ for $x\ge \sqrt 3$, $$0\lt h(x)\le 1\iff 0\lt \frac{x^3-3x-2}{x^3-3x+2}\le 1\iff 0\lt x^3-3x-2\le x^3-3x+2$$ $$\iff 0\lt x^3-3x-2=(x-2)(x+1)^2\iff x\in (2,\infty)$$ Therefore, in Case 5, we have $$x\in\left(\sqrt 5,\frac 73\right)\tag5$$

Case 6 : $x\gt 0$ and $1\lt g(x)\lt f(x)$ and $0\lt h(x)\le 1$

Since $x\gt 0$, we have $g(x)=x^2-3x+2$.

Case 6-1 : For $x\in\left(0,\sqrt 3\right)$, we have $f(x)=-(x^2-3)-2=-x^2+1$ and $$1\lt g(x)\lt f(x)\iff 1\lt x^2-3x+2\lt -x^2+1$$ There are no such $x$.

Case 6-2 : For $x\in [\sqrt 3,\infty)$, we have $$f(x)=(x^2-3)-2=x^2-5,\quad h(x)=\frac{x^3-3x-2}{x^3-3x+2}$$

$$1\lt g(x)\lt f(x)\iff 1\lt x^2-3x+2\lt x^2-5\iff x\in\left(\frac{3+\sqrt 5}{2},\infty\right)$$ $$0\lt h(x)\le 1\iff x\in (2,\infty)$$

Therefore, in Case 6, we have $$x\in\left(\frac{3+\sqrt 5}{2},\infty\right)\tag6$$

Case 7 : $x\gt 0$ and $0\lt g(x)\lt f(x)\lt 1$ and $h(x)\ge 1$

Since $x\gt 0$, we have $g(x)=x^2-3x+2$.

Case 7-1 : For $x\in\left(0,\frac 23\right)$, we have $$f(x)=-(x^2-3)-2=-x^2+1,\quad h(x)=\frac{x^3-(3x+2)}{x^3-(-(3x-2))}=\frac{x^3-3x-2}{x^3+3x-2}$$ $$0\lt g(x)\lt f(x)\lt 1\iff 0\lt x^2-3x+2\lt -x^2+1\lt 1\iff x\in\left(\frac 12,1\right)$$ Since $j(x)=x^3+3x-2$ is increasing with $j(\beta)=0$ where $\frac 12\lt \beta\lt \frac 23$, we have

For $x\lt\beta$ where $x^3+3x-2\lt 0$, $$h(x)\ge 1\iff \frac{x^3-3x-2}{x^3+3x-2}\ge 1\iff x^3-3x-2\le x^3+3x-2\iff x\ge 0$$
For $x\gt\beta$ where $x^3+3x-2\gt 0$, $$h(x)\ge 1\iff \frac{x^3-3x-2}{x^3+3x-2}\ge 1\iff x^3-3x-2\ge x^3+3x-2\iff x\le 0$$

from which we have $$h(x)\ge 1\iff 0\lt x\lt\beta$$

So, in this case, we have $$x\in\left(\frac 12,\beta\right)$$ where $\beta\approx 0.596$ is the only real root of $x^3+3x-2$.

Case 7-2 : For $x\in \left[\frac 23,\infty\right)$, we have $$h(x)=\frac{x^3-(3x+2)}{x^3-(3x-2)}=\frac{x^3-3x-2}{x^3-3x+2}$$ Since $x^3-3x+2=(x+2)(x-1)^2\gt 0$ for $x\in \left[\frac 23,\infty\right)$ with $x\not=1$, $$h(x)\ge 1\iff \frac{x^3-3x-2}{x^3-3x+2}\ge 1\iff x^3-3x-2\ge x^3-3x+2$$ There are no such $x$.

Therefore, in Case 7, we have $$x\in\left(\frac 12,\beta\right)\tag7$$ where $\beta\approx 0.596$ is the only real root of $x^3+3x-2$.

Hence, the answer is $(4)\cup (5)\cup (6)\cup (7)$, i.e. $$\small\color{red}{x\in\left(-\sqrt 6,-\frac 73\right)\cup \left(-1,-\frac 12\right)\cup \left(\frac 12,\sqrt[3]{1+\sqrt 2}-\frac{1}{\sqrt[3]{1+\sqrt 2}}\right)\cup \left(\sqrt 5,\frac 73\right)\cup \left(\frac{3+\sqrt 5}{2},\infty\right)}$$ where $\sqrt[3]{1+\sqrt 2}-\frac{1}{\sqrt[3]{1+\sqrt 2}}=\beta\approx 0.596$ is the only real root of $x^3+3x-2$.

(From Sid's comment, we can find $\beta$ by setting $x=y-\frac 1y$ for $x^3+3x-2=0$ to have $(y^3)^2-2y^3-1=0$ which is a quadratic equation on $y^3$.)

Finally, let us prove Lemma 1 and Lemma 2.

Lemma 1 : If $x\lt 0$, then $h(x)\lt 1$.

Proof :

$$h(x)\ge 1\iff \frac{x^3-|3x+2|}{x^3-|3x-2|}\ge 1$$

For $x\in\left(-\infty,-\frac 23\right)$, since $x^3+3x-2\lt 0$, $$h(x)\ge 1\iff \frac{x^3+3x+2}{x^3+3x-2}\ge 1\iff x^3+3x+2\le x^3+3x-2$$which does not hold.
For $x\in\left[-\frac 23,\beta\right)$ where $\beta$ is the only real root of $x^3+3x-2$, since $x^3+3x-2\lt 0$, $$h(x)\ge 1\iff\frac{x^3-3x-2}{x^3+3x-2}\ge 1\iff x^3-3x-2\le x^3+3x-2\iff x\ge 0$$
For $x\in\left(\beta,\frac 23\right)$, since $x^3+3x-2\gt 0$, $$h(x)\ge 1\iff \frac{x^3-3x-2}{x^3+3x-2}\ge 1\iff x^3-3x-2\ge x^3+3x-2\iff x\le 0$$
For $x\in\left[\frac 23,\infty\right)$, since $x^3-3x+2=(x+2)(x-1)^2\gt 0$ with $x\not=1$, $$h(x)\ge 1\iff \frac{x^3-3x-2}{x^3-3x+2}\ge 1\iff x^3-3x-2\ge x^3-3x+2$$which does not hold.

So, the claim follows from that $h(x)\ge 1\implies x\ge 0$. $\quad\blacksquare$

Lemma 2 : There are no $x$ such that $f(x)\gt 1$ and $g(x)\gt f(x)$.

Proof :

$$f(x)\gt 1\iff |x^2-3|-2\gt 1\iff |x^2-3|\gt 3$$ $$\iff x^2-3\lt -3\quad\text{or}\quad x^2-3\gt 3\iff x\in (-\infty,-\sqrt 6)\cup (\sqrt 6,\infty)$$ On the other hand, $$g(x)\gt f(x)\iff x^2-3|x|+2\gt |x^2-3|-2$$

For $x\in (-\infty,-\sqrt 6)$, $$g(x)\gt f(x)\iff x^2-3(-x)+2\gt (x^2-3)-2\iff x\in \left(-\frac 73,\infty\right)$$
For $x\in (\sqrt 6,\infty)$, $$g(x)\gt f(x)\iff x^2-3x+2\gt (x^2-3)-2\iff x\in\left(-\infty,\frac 73\right)$$

The claim follows from that $\frac 73\lt \sqrt 6$. $\quad\blacksquare$

Best Answer

Related Solutions

Imaginary Numbers – Exploring x^c = -x

Inequality – Solving Logarithmic Inequality in Two Hours

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