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How do I take natural logarithm of the following?

$A - (Be^{-xy})$

Newb
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user101066
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  • Is your problem $A-(Be^{-xy})$ or $(A-B)e^{-xy}$? – Newb Oct 23 '13 at 13:56
  • It is A - (Be^-xy) – user101066 Oct 23 '13 at 13:58
  • Is this part of an equation? Like the equation $A - Be^{-xy} = 0$? – amWhy Oct 23 '13 at 13:59
  • The question is totally unclear. I have flagged it until this is made clearer. – Ahaan S. Rungta Oct 23 '13 at 14:00
  • Like this: $\ln(A-Be^{-xy}).$ There's really nothing else we can do with that, though. – Cameron Buie Oct 23 '13 at 14:03
  • @CameronBuie Yeah, that's what I was thinking. See my answer (you beat me by 27 seconds). – Ahaan S. Rungta Oct 23 '13 at 14:03
  • @AhaanRungta Flagging the question four minutes after it posted, without waiting just a tad longer for the OP to have a chance to clarify or respond to questions, is what I consider "jumping the gun." Give an OP the chance to respond! – amWhy Oct 23 '13 at 14:14
  • @amWhy The OP saw our comments and refused to clarify, which is why I thought the clarification would never be made. But it seems you understood his/her question and have answered it and that's what matters! ;) – Ahaan S. Rungta Oct 23 '13 at 14:16
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    @AhaanRungta: Sometimes, new users take a bit longer to reply, or are unaware that they can edit their question. There is a difference between "refused to clarify" and "has not clarified yet." Try to be patient with them. – Cameron Buie Oct 23 '13 at 14:24
  • Sorry! I'll try to keep that in mind in future. Thanks. :) – Ahaan S. Rungta Oct 23 '13 at 14:33

2 Answers2

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If you have the equation shown below on the left hand side of $\iff$, we can do the following: $$A - Be^{-xy} = 0 \iff Be^{-xy} = A$$ Now we can take the natural logarithm of each side:

$$\begin{align} \ln\left(Be^{-xy}\right) & = \ln A \\ \\ \iff \ln B + \ln\left(e^{-xy}\right) & = \ln A \\ \\ \iff -xy & = \ln A - \ln B = \ln \left(\dfrac AB\right)\end{align}$$

Otherwise, if you are simply trying to take the logarithm of an expression, there's nothing you can really do to simplify the expression, so you'd have nothing simpler than $$\ln\left(A - B^{-xy}\right)$$

amWhy
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Until the question is made clearer, here is what I have to say. This is nothing you can do to simplify this further because it is $A-Be^{-xy}$, not $Be^{-xy}$ or $\frac{A}{Be^{-xy}}$. You cannot use logarithmic properties with such a form.

Ahaan S. Rungta
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