# Simple mathematical problem

1. Feb 5, 2014

### Mechatron

Is this equation equal to:

(e^(hf/kT)) - 1

or

e^( (hf/kT) - 1 )

http://s29.postimg.org/le6iqy3rb/exp.png [Broken]

Last edited by a moderator: May 6, 2017
2. Feb 5, 2014

The former.

3. Feb 5, 2014

### Staff: Mentor

First off, what you wrote is NOT an equation. An equation always has an = symbol in it.

The image in the link is [exp(hf/kT) - 1].

What you have written is ambiguous, as what you probably meant is this:
$$e^{\frac{hf}{kT} - 1}$$

What you actually wrote, though, is this:
$$e^{\frac{hf}{k}T - 1}$$

The brackets - [] - around the entire expression are unnecessary.

Last edited by a moderator: May 6, 2017
4. Feb 5, 2014

### Staff: Mentor

Why did you put the -1 in the exponential? The parenthesis limit the argument of exp to hf/kT.

My guess is that Mechatron did not write that himself, but saw it in a book. It's most probably related to the Planck distribution (blackbody radiation). As economicsnerd said, the correct reading is
$$e^{\beta h f} - 1 \mbox{ where } \beta = \frac{1}{kT}$$
The additional bracket [] might be there because it is part of a greater equation.

5. Feb 15, 2014

### Sidney

Im sorry to be off topic but I'm using a i device and i can't see (what i think to be) mathematical symbols that are in this thread... They appear as dollar signs and other randoms, i was wondering if maybe I'm short of additional download or setting adjustments. All help is highly appreciated

6. Feb 15, 2014

### Staff: Mentor

The posted image, which doesn't have the -1 term, doesn't match the expressions in the first post. In the first post Mechatron asks about these expressions:
(e^(hf/kT)) - 1

and

e^( (hf/kT) - 1 )

In any case, this is moot, as Mechatron has been banned from PF.