Out of curiosity... why does this = 1?

Question

anonymous · Answer

it doesn't

anonymous · Answer

what it the base of your logarithm?

parthkohli · Answer

I'd be the Google. 

Did you mean: $\log_5 5 = 1$?

anonymous · Answer

@satellite73 The base is 5

anonymous · Answer

so maybe 
$$5^{\log_5(5)}$$?

parthkohli · Answer

So is it $5^{\log_5 5} = 1$?
That's 5; not 1.

anonymous · Answer

or maybe 
$$5^{\log_5(1)}$$

parthkohli · Answer

Though it is $1$ in $\log_5 5 = 1$.

anonymous · Answer

$$a ^{\log_{a}12} = 12$$

parthkohli · Answer

Yup :)

anonymous · Answer

So doesnt that mean $$a ^{\log_{a} } = 1 $$ ?

parthkohli · Answer

No, not really. How does the logarithm have a base but not the top part?

anonymous · Answer

Oh... right. Wow, duh haha.
I don't really understand how to comprehend the ones where the log is the exponent...

parthkohli · Answer

It's pretty easy once you get the analogy!

$\log_a b$ is another way to say, “$a$ to the what power is $b$?”

parthkohli · Answer

Let me get it a little more right.

parthkohli · Answer

If you have two numbers $a$ and $b$, then $\log _a b$ is the answer to 'what' in, “$a$ to the what power is $b$?”.

anonymous · Answer

not to butt in, but this question 
$$a ^{\log_{a} } = 1$$ doesn't make any sense
log is a function, you have to take the log of something

parthkohli · Answer

So if you ponder a bit more, you see that the thing $a^{\log_a 12} = 12$ is just a meta-thingy.

anonymous · Answer

in other words, does not make sense to write 
$$\log_a$$ without some variable of number after 
$$\log_a(x)$$ or $$\log_a(75)$$ or something.

parthkohli · Answer

If you raise $a$ to the power that you have to raise $a$ to, for getting $12$, you'd get 12.

parthkohli · Answer

It's a little confusing but you'd get the point.

anonymous · Answer

@Parthkohli :O I'm going to have to think about this haha

@satellite73 Yes, thank you :)

parthkohli · Answer

@1210: You have to raise $a$ to SOME power, to get $12$. The "SOME" is $\log_a 12$. Now if you raise $a$ to "SOME" power, wouldn't you get $12$?

anonymous · Answer

what you need to know cold is how to switch form logarithmic form to exponential form, which really means understand what a log is 
$$\log_b(x)=y\iff b^y=x$$ so for example 
$$\log_5(25)=y\iff 5^y=25$$ etc

anonymous · Answer

in other words, at least as an introduction, a logarithm is an exponent
if you remember that a log is an exponent, it makes all those weird rules seem rather obvious