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I'm going to rephrase this because I seem to be confusing people.

If I have $a+b=c$

I can say $\ln a+\ln b=\ln c$

But if I have $\ln a+\ln b=\ln c$

I can't say $a+b=c$

Why?

Marc van Leeuwen
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Korgan Rivera
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    The only reason your last example is nonsense is in the last line... $\ln a + \ln b = \ln c$ does NOT imply $a + b = c$. At best $\ln a + \ln b = \ln c$ implies $\exp(\ln a + \ln b) = c$. – Shaun Ault Sep 14 '13 at 16:07
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    Your line three is in error. You can say that if $a+b=c$, then $\ln(a+b)=\ln c$, provided that both $a+b$ and $c$ are positive (i.e. in the domain of $\ln$). But $\ln a+\ln b=\ln(ab)$, and different from $\ln a+\ln b$. OTOH, if $a,b>0$, then the question in the TITLE is answered in the affirmative, because $\ln x$ is injective. This implication is precisely the definition of injectivity as is being explained in several answers. – Jyrki Lahtonen Sep 14 '13 at 16:15
  • $\ln 1 + \ln 1 = 0 \neq \ln 2 $ – user92797 Sep 14 '13 at 16:17
  • @JyrkiLahtonen Yes! I see it now. It seems so obvious now. Thank you! – Korgan Rivera Sep 14 '13 at 16:18
  • Good! ${}{}{}{}$ – Jyrki Lahtonen Sep 14 '13 at 16:20
  • The title question does not correspond at all to the body question. The answer to the title question is yes. Just apply $\exp$ to both sides. – Marc van Leeuwen Sep 14 '13 at 16:34

3 Answers3

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No, if $a+b=c$ you can say $\log(a+b)=\log c$. What you might be thinking is, if $ab=c$ then $\log ab=\log a+\log b=\log c$, and conversely. As I had written, consider $$a^2+b^2=c^2$$ Why should it follow that $a+b=c$?

Pedro
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If $a = b$ then $f(a) = f(b)$ for any function with $a$ in its domain. The converse is not true. That is, if $f(a) = f(b)$, then $a$ is not necessarily equal to $b$. If $f : X \to Y$ is such that $f(a) = f(b)$ implies $a = b$, $f$ is said to be injective (or $1-1$). Injectivity is equivalent to the existence of an inverse function $f^{-1} : f(X) \to X$ which satisfies $(f^{-1}\circ f)(x) = x$ for every $x \in X$ and $(f\circ f^{-1})(y) = y$ for every $y \in f(X)$.

In the example you're considering, the exponential function is the inverse of the natural logarithm. That is $\exp(\ln(x)) = x$ for every $x \in (0, \infty)$ and $\ln(\exp(y)) = y$ for every $y \in \mathbb{R}$. Therefore, if you have $\ln(a) = \ln(b)$, by the first sentence in this answer, I can apply the exponential function to both sides which gives $\exp(\ln(a)) = \exp(\ln(b))$. Now $\exp(\ln(a)) = a$ and $\exp(\ln(b)) = b$, so we obtain $a = b$.

Added to address the edit: If $a + b = c$ you cannot say that $\ln(a) + \ln(b) = \ln(c)$. For example $1 + 1 = 2$ but $\ln(1) + \ln(1) = 0 \neq \ln(2)$.

Michael Albanese
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If you have $a=b$ you can apply any function $f$ and get $f(a)=f(b)$.

A function $f$ does need to be well defined meaning if $x=y$ then $f(x)=f(y)$.

Dominic Michaelis
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  • I know that part. What I'm asking is, is it true the other way around? If $f(x)=f(y)$, is $x=y$? And if not, why not? – Korgan Rivera Sep 14 '13 at 16:06
  • @KorganRivera oh in the body of your question it seems to be the other way around. In general if $f(x)=f(y)$ then it doesn't follow that $x=y$ this is only true for injective functions (which are defined that way). As the logarithm is injective you know that $\log(x)=\log(y)$ implies that $x=y$ – Dominic Michaelis Sep 14 '13 at 16:08