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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 $\ln(x+y)$. – 2017-01-01
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    This may help: http://math.stackexchange.com/questions/1538477/log-of-summation-expression – 2017-01-01
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    can $\ln({ae^b + ce^d})$ be simplified? – 2017-01-01
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    if $b$ is different from $d$ I can't see how, in general – 2017-01-01
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    The question is "*Solve $3e^y + 5e^{-y} = 16$*" – 2017-01-01
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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 "$+$". – 2017-01-01
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    In the particular case you have just written you can get the $y$ out, you just have to factorize $e^y$ – 2017-01-01
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    Without logging? – 2017-01-01
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    @Tobi See the edit to my question. – 2017-01-01
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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? – 2017-01-01
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    just as @NoahSchweber has described in a more general case – 2017-01-01
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    @NoahSchweber :$ 3e^y + 5e^{-y} = 3e^y + 5e^{y}\,e^{-1}= (3 + 5\,e^{-1})e^{y}$ – 2017-01-01
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    oh sorry, my mistake – 2017-01-01
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    @kdow **No, it doesn't**: exponents *add* when you multiply the terms. $e^ye^{-1}=e^{y-1}$, not $e^{-y}$. – 2017-01-01
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    XD, very dumb mistake. I'll get it right – 2017-01-01

2 Answers 2

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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?

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    Thank you, in my working out I'd written "$3e^{2y} + 5 = 16e^y$" but forgot $e^{2y}$ was $(e^y)^2$ – 2017-01-01
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    The last equation is a simple quadratic. – 2017-01-02
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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 $x=0$). – 2017-01-02
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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.