How is log (1/2) = - log (2)?

Question

How is log (1/2) = - log (2)?

Answer 1

log(1/2) = log(1) - log(2) = 0 - log(2) = -log(2)

Answer 2

I think many people may get formulaic with this answer and will likely give correct responses. If you are looking for an understanding however perhaps it is better to think of it in terms of exactly what a “logarithm” is.

Log x is the number you must raise the base to in order to get x (base may be 10 or the number e, it doesn’t matter in this case): i.e. log x = a if and only if (Base)^a = x.

Since ½ and 2 are reciprocals, each is the other raised to the power of -1: x^(-1) = 1/x

Hence, if you know the log of a number, then the log of its reciprocal will always be the negative of that, because the result of raising a number to some power, and raising that same number to the negative of that power will give two answers which are reciprocals of each other.

If you want to see it formula wise:

If (base)^a = x

Then (base)^(-a) = 1/x

Hence, by the “if and only if” stated above, this means

Log x = a

And

Log (1/x) = -a

So

Log x = - Log (1/x) or -log(x) = log(1/x)

Of course this won’t work for x = 0, (Why? You can’t raise a non-zero Base to a finite power and get zero, but that’s another story), but hopefully the main points are clear.

Hope this helps.

Answer 3

In the properties of logarithms: multiplication is kind of like addition, and division is kind of like subtraction. So in a way, log(1/2) = log(1) - log(2). Since log(1)=0, then this simplifies to 0 - log(2) = -log(2). Hope this helps.

Answer 4

log m/n = log m - log n
If 10^m = n then log n = m
Since 10^0 = 1 so log 1 = 0

log 2/1 = log 2 - log 1 = log 2 - 0 = log 2
log 1/2 = log 1 - log 2 = 0 - log 2 = -log 2

Answer 5

You just applied these two logarithmic rules and you
will prove that log(1/2) = -log(2).

Rule1 : log (a/b) = log (b/a)^-1
Rule 2 : log a^n = nlog a


log(1/2) = log(2/1)^ -1 = log (2)^ -1 (Rule 1)

log (2)^ -1 = -1log(2) = -log(2) (Rule 2)


So, log(1/2) = -log(2)

Answer 6

log(1/2) = log 1 - log 2
log(1/2) = 0 - log 2
log(1/2) = - log 2

Answer 7

log(1/2)=log(2^-1) so it makes - log(2)

Answer 8

use log(a.b) = log(a)+log(b) from which you can deduce log(a/b) = log(a)-log(b)

since log((a/b) x b) = log (a) (just reducing the part inside the parenthesis)
using above property, this also equal to
log((a/b) x b) = log(a/b) + log(b) ,so we have log(a) = log(a/b) + log(b) which is the same as saying log(a/b) = log(a) - log(b)

now using this fact

log(1/2) = log(1) - log(2)
also you should know log(1) = 0 (this is due to the definition of log(x) which is the integral of 1/s from 1 to x

I wouldn't be able to demonstrate why log(a x b) = log(a) + log(b), without reverting to the fact that taking a logarithm is the inverse of exponentiating so that exponentiating both sides of
log(a.b)=log(a)+log(b) we can check
left side : exp(log(a.b)) = a.b
right side exp(log(a) + log(b)) = exp(log(a)) times exp(log(b)) = a times b

Answer 9

1/2 = 2^-1 , with log(2)^-1 the -1 power goes out front or -1log(2)

Answer 10

Okay, so don't listen to the guy that said there's only 3 bases lol. Obviously he hasn't played baseball OR fooled around with anyone hahaha. But I'm assuming your friend probably just made that up. There's no actual term for "3 1/2 base". Since 3rd base is oral, and 4th is all the way. I'm assuming maybe he fingered her or something? It just wasn't technically intercourse since he didn't use his penis lol. That's my best guess haha