log(-1) = ??

If x^2 = 1 then, x = -1, 1

In logarithm,

log(x^2) = log(1)

or, log((-1)^2) = 0

or, 2 * log(-1) = 0

so, log(-1) = 0

but we know that log(-1) is invalid. What is the problem in this equation??

#Logarithms

Easy Math Editor

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Comments

Look at the reverse process: $10^0=1≠-1$

Trevor B. - 7 years, 4 months ago

now how can i prove that x = -1 form that equation using logarithm?

Fahim Rahman - 7 years, 4 months ago

While $x=-1$ is a solution to the equation $x^2=1,$ it is not the $\textit{principle}$ root. This is the positive square root. By convention, the square root of $1$ is $1,$ not $-1,$ because $1$ is the principle square root.

The mistake is in the fourth line of the proof, where you say that the $\sqrt{1}=-1.$

Trevor B. - 7 years, 3 months ago

log x^2=log 1 10^log 1=x^2 {using property a^logN(at base a)=N} 1=x^2 we are again at the same solution , x=(1,-1). But in the given proof you cannot use the value of x =-1 in the fourth step right like Trevor says.

Raven Herd - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$