If $f(x) = ax +b$,
is $\ln{f(x)} \equiv \ln(ax+b)$ or $\ln{ax}+\ln{b}$?
If it is $\ln(ax+b)$ how should I go about splitting or simplifying $\ln{(ae^{c}+b)}$?
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The property of the logarithm is that, for $x,y>0$, $\ln(xy)=\ln x+\ln y$. You can't say much about $\ln(x+y)$. – egreg Jan 01 '17 at 23:25
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This may help: http://math.stackexchange.com/questions/1538477/log-of-summation-expression – calm-tedesco Jan 01 '17 at 23:26
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can $\ln({ae^b + ce^d})$ be simplified? – Tobi Jan 01 '17 at 23:28
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if $b$ is different from $d$ I can't see how, in general – calm-tedesco Jan 01 '17 at 23:30
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The question is "Solve $3e^y + 5e^{-y} = 16$" – Tobi Jan 01 '17 at 23:32
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@Tobi It can be simplified a bit. Say $b\le d$; then we can factor out $e^b$, and rewrite the expression as $b+\ln(a+ce^{d-b})$. But this isn't much better. If $b=d$, we get $b+\ln(a+c)$, which is noticeably better. But we can't get rid of the "$+$". – Noah Schweber Jan 01 '17 at 23:32
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In the particular case you have just written you can get the $y$ out, you just have to factorize $e^y$ – calm-tedesco Jan 01 '17 at 23:34
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Without logging? – Tobi Jan 01 '17 at 23:34
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@Tobi See the edit to my question. – Noah Schweber Jan 01 '17 at 23:35
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@kdow I don't think that's right - note that the exponent of the second term is "$-y$". Or, what do you have in mind? – Noah Schweber Jan 01 '17 at 23:35
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just as @NoahSchweber has described in a more general case – calm-tedesco Jan 01 '17 at 23:36
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@NoahSchweber :$ 3e^y + 5e^{-y} = 3e^y + 5e^{y},e^{-1}= (3 + 5,e^{-1})e^{y}$ – calm-tedesco Jan 01 '17 at 23:38
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oh sorry, my mistake – calm-tedesco Jan 01 '17 at 23:39
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@kdow No, it doesn't: exponents add when you multiply the terms. $e^ye^{-1}=e^{y-1}$, not $e^{-y}$. – Noah Schweber Jan 01 '17 at 23:39
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XD, very dumb mistake. I'll get it right – calm-tedesco Jan 01 '17 at 23:39
2 Answers
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$\ln(f(x))$ is $\ln(ax+b)$; you plug "$f(x)$" directly into "$\ln x$." And there is no real way to simplify an expression of the form "$\ln(A+B)$". For instance, it's certainly not the same as $\ln(A)+\ln(B)$ - take for example $A=B=1$, then $\ln(A) + \ln(B) = 0 + 0 = 0$ but $\ln(A+B)=\ln(2)>0$.
Is there a reason you believe you can simplify the expression you mention?
In the comments you mention that the question you're trying to answer is "Solve $3e^y+5e^{-y}=16$."
A natural instinct is to use logs, but as you've seen already that doesn't work - the problem you run into is that $\ln(A+B)$ is unsimplifiable.
So what else can you do? Well, one good idea is to try to first solve for "$e^y$". That is, find out what "$x$" is in the equation $$3x+5x^{-1}=16.$$ That's still not great, but there's something you can do to this now to get it into a form you know how to solve - do you see how?
Blencer
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Noah Schweber
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Thank you, in my working out I'd written "$3e^{2y} + 5 = 16e^y$" but forgot $e^{2y}$ was $(e^y)^2$ – Tobi Jan 01 '17 at 23:37
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One thing other answers haven't mentioned is it can sometimes be useful to rewrite $\ln\left( ax+b\right)$ as $\ln b + \ln\left( 1+\frac{ax}{b}\right)$ for small $x$ or as $\ln x + \ln a + \ln\left( 1+\frac{b}{ax}\right)$ for large $x$, because of the approximation $\ln\left( 1+z\right)\approx z$. These expressions are correct for any $x$ (except that the large-$x$ case breaks down for $x=0$). – J.G. Jan 02 '17 at 08:22
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If $f(x) = ax+b$, then $\ln f(x) = \ln(ax+b)$. You cannot simplify $\ln(ae^c+b)$ more, perhaps except in special cases.
Eff
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