why is loga a = 1 false if a is a real number?

Question

ganeshie8 · Answer

who says it is false ?

freckles · Answer

there are certain restrictions to a

ganeshie8 · Answer

oh right, it need not be true always...

anonymous · Answer

true or false: $$\log_{a} a=1$$ if $$a$$

anonymous · Answer

*if a is a real number

anonymous · Answer

and the answer is false...

ganeshie8 · Answer

$$\log_a a = 1 \iff  a = a^1$$

see any real numbers, $a$, that might break above equality ?

poopsiedoodle · Answer

o well dang I was about to do a big ol' explanation there because I thought you made a typo but now I have no idea :<

jim_thompson5910 · Answer

The domain of log(x) is x > 0

freckles · Answer

$$\log_a(a)=\frac{\ln(a)}{\ln(a)}$$

freckles · Answer

consider the domain of that expression

ganeshie8 · Answer

it seems $a=0$ may be an issue

anonymous · Answer

so it doesn't work whenever a is equal to 0...

freckles · Answer

$$\frac{\ln(a)}{\ln(a)} \ \text{ we want } \ln(a) \neq 0 \text{ and we want } a>0$$

poopsiedoodle · Answer

why wouldn't it work if a is 0 tho? because 0^0 still = 1

jim_thompson5910 · Answer

a = 1 doesn't work either (plug a = 1 into the equation @freckles wrote)

jim_thompson5910 · Answer

0^x = 0 where x is nonzero
x^0 = 1 where x is nonzero

0^0 is one of many indeterminate forms

freckles · Answer

the intersections of the set of number a such that ln(a) not equal to 0 and a>0
is...

ganeshie8 · Answer

Yes, $\log_a a=1$ holds for all real $a$ except $\{0,1\}$

freckles · Answer

I don't think it holds for the negative numbers

ganeshie8 · Answer

why not

freckles · Answer

because the domain of ln(x) is x>0

freckles · Answer

i guess it depends on what class this is

anonymous · Answer

I still don't quite get it... because $$1=1^{1}$$ and $$0=0^{1}$$

ganeshie8 · Answer

that is a good question, 
i have the same question too

anonymous · Answer

this is calculus 1 class

ganeshie8 · Answer

$\log_0 0$ is not well defined because, it could be any thing. you may also say : 
$$\log_0 0 = 2 $$

ganeshie8 · Answer

or
$\log_0 0 = 3 $

or anything you wish because $0=0^n$ for all finite nonzero $n$

ganeshie8 · Answer

i think, thats the reason we let $\log_0 0$ be undefined

anonymous · Answer

ok, thank you everyone! I think I got it...

jim_thompson5910 · Answer

Using @ganeshie8 's reasoning, $\Large \log_{1}(1)$ is also poorly defined


$\Large \log_{1}(1) = x$ --> $1^x = 1$ --> $1 = 1$ where x is any real number

ganeshie8 · Answer

Exactly! I think those are the only two bad real numbers that `log` can't deal with