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I'm not good at Math and please don't blame me if this question is too obvious to you. There is a formula about logarithmic function.

$\log_ab^x = x\times\log_ab$

Here is a problem:

$\log_5x^2 = -2$

If I use the formula, I got:

$2\times log_5x=-2$

But it seems this is not correct. In the problem, $x$ could be any number including negtive numbers. After using the formula, $x$` can't be negative. So is this formula wrong or am I doing wrong somewhere?

Just a learner
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  • Your initial formula is only valid for positive $b$. For $b \le 0$ the right-hand side is not defined. – Martin R Jun 01 '19 at 17:29
  • That's fine if $b$ is positive. – Angina Seng Jun 01 '19 at 17:29
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    If $n$ is odd, then $\log_a \left(b^n\right) = n\log_a (b)$ for $b > 0$, but when $n$ is even, then $\log_a \left(b^n\right) = n\log_a \vert b\vert$ for $b \neq 0$. Notice the absolute value. So, in the last step you've shown, you should actually have $2\log_5 \vert x\vert = -2$. – KM101 Jun 01 '19 at 17:30
  • For clarification, this is since a number raised to an odd power will retain its sign, so $\log_a \left(b^n\right)$ (for odd $n$) will only be defined for positive $b^n$, and hence, positive $b$. But when a non-zero number is raised to an even power, it will always turn out positive. Hence, $\log_a \left(b^n\right)$ (for even $n$) is defined for all non-zero $n$. (This is why the absolute value is used.) – KM101 Jun 01 '19 at 17:37

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You are right. In the original equation, $x$ can be negative. Writing $2\log_5x=-2$ will yield only the positive root. A better way to solve the equation is to use the definition of logarithm directly.

\begin{align*} \log_5x^2&=-2\\ x^2&=5^{-2}=\frac1{25}\\ x&=\pm\frac15 \end{align*}

CY Aries
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