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# 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??

Note by Fahim Rahman
3 years, 11 months ago

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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.

- 3 years, 8 months ago

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

- 3 years, 11 months ago

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

- 3 years, 11 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 $$\sqrt{1}=-1.$$

- 3 years, 11 months ago