3

In the equation below,

Solve for $x$: $\log\left(x^2\right)=4$

Here, I think that the answer is going to be $100$ and $-100$. Because if we insert $-100$, $x^2$ is still positive and thereby, it doesn't violate the rule that we can only insert positive values in logarithm.

But another counter point I have is that, we can write the expression as,

2log(x)=4

Or, log(x)=2

Or, x=10^2

In this way, I get the only positive value. Which one of them is correct? And if process 1 is, what are the flaws in the second?

gt6989b
  • 54,422
Habib
  • 145

3 Answers3

3

If it's $\log_{10}$, use the definition before the first step, because the argument of $\log$ is postive, and then solve for $x$: $$ x^2 = 10^4\\ (x-100)(x+100)=0 $$

Alex
  • 19,262
3

The fact is that $\ \log\left(x^2\right) = 2\log(x)\ $ if and only if $\ x > 0.$

If $x < 0$ then $\log\left(x^2\right)$ is defined but $2\log(x)$ is not, so the equation is false.

The method using the equation $\log\left(x^2\right) = 2\log(x)$ will find a positive solution if one exists. But if there is a negative solution, that method cannot find it.

A more general statement is $$ \log\left(x^2\right) = 2\log\left(\lvert x\rvert\right), $$ which is true whenever $x \neq 0.$ Using that fact, you can find both solutions.

David K
  • 98,388
  • So, to be clear, log(x^n) = nlog(।x।) if n is even and, if n is an odd number, it is log(x^n)= nlog(x). Ami I right? – Habib Aug 23 '20 at 17:22
  • 1
    Yes. It is because $x^n = \lvert x\rvert^n$ when $n$ is an even integer. And you are correct, the equation with the absolute value is wrong if $n$ is odd, because then $x^n$ has the same sign as $x.$ – David K Aug 23 '20 at 17:27
1

Use the following two properties:

$u=\log_b(v)$ if and only if $v=b^u$

$u^2=a$ if and only if $u=\sqrt{a}$ or $u=-\sqrt{a}$

Ben W
  • 5,236