Why are 1/log_a(e) and log_e(a) reciprocals?

[alijan kk]

Homework Statement

1/log_a(e) = log_e(a)

Homework Equations

The Attempt at a Solution

how they are reciprocals of each other ? is their any longer but intuative way to show this result

 

[phyzguy]

I think you mean:
$$\frac{1}{\log_a(e)} = \log_e(a)$$

 

[alijan kk]

I think you mean:
$$\frac{1}{\log_a(e)} = \log_e(a)$$

yes sir i mean that

 

[Dick]

how they are reciprocals of each other ? is their any longer but intuative way to show this result

$a^{\log_a(e)}=e$, right? Just take $\log_e$ of both sides.

 

[Mark44]

There's an important part missing from your problem statement:

Homework Statement

Prove that[/B] 1/log_a(e) = log_e(a)

Along the lines of Dick's hint are these relationships:
$y = \log_a(x) \Leftrightarrow x = a^y$
I.e., the two equations are equivalent: any pair of values (x, y) that satisfies the first equation also satisfies the second equation, and vice versa.

 

[scottdave]

There's an important part missing from your problem statement:

Along the lines of Dick's hint are these relationships:
$y = \log_a(x) \Leftrightarrow x = a^y$
I.e., the two equations are equivalent: any pair of values (x, y) that satisfies the first equation also satisfies the second equation, and vice versa.

Here is a way that I like to remember it. When I see $y = \log_a(x)$ and want to convert it to something like $x = a^y,$ I use this to help remember.

Think of log base 10. So we have $y = \log_{10}(1000)$. This is pretty easy, there are 3 zeros and the answer is y = 3. The log is equivalent to the exponent.
We have 10³ = 1000.

 

[Mark44]

The log is equivalent to the exponent.

Not only that -- a logarithm is by definition an exponent. Specifically, $\log_a(x)$ represents the exponent on a that produces x.
Using your example, $\log_{10}(1000) = 3$ because $10^3 = 1000$.

 

[Gene Naden]

I think the relation in the problem statement is true for all positive real a and e. I found this rather difficult to prove. I had to go to a spreadsheet and plug in some values to convince myself it was true. My proof starts with the statement "a=a." Then I substitute an exponential for a on the left hand side. I think I am not supposed to give the full solution here. Good luck! :)

 

[Mark44]

I think the relation in the problem statement is true for all positive real a and e.

It wasn't stated in the first post, but the equation is an identity. Yes, it is true for all a > 0, and e is "the natural number," approximately 2.718.

I found this rather difficult to prove.

It's not difficult to prove. Let $y = \log_a(e)$. This is equivalent to the equation $e = a^y$. Substitute for e in the expression on the left side of the original equation, $\frac 1 {log_a(e)}$, and within a couple of steps you end up with the expression on the right side.

 

[scottdave]

You can convert a log of one base to another base. I'm on my phone so I'm not sure if I can do LateX correctly though. If you have log(x) in base b. You can calculate it by LN(x) / LN(b), where LN is the natural log (log base e)

 

[scottdave]

So using that, substitute LN(e) for 1.

 

[scottdave]

Oops, I meant log_a(a) = 1