# Calculating $\ln(1+\sqrt3)$

I distributed the natural logarithm and got $(0 + 0.549)$ [placing the values in a calculator]. However, the answer key states that the answer is $1.0051$.

Where did I go wrong?

• Four answers and so far I'm (I think?) the only one who's up-voted the question. – Michael Hardy Aug 10 '14 at 21:16

The logarithm is not linear. That is, typically $\ln(x+y)\neq\ln(x)+\ln(y)$.

• Not just "typically". $\ln(x) + \ln(y) = \ln(xy)$, so $\ln(x+y) \ne \ln(x)+\ln(y)$ unless $x+y=xy$. For each $y \ne 1$, there is exactly one real $x$ which makes $\ln(x+y)= \ln(x)+\ln(y)$. For $0 < y < 1$ there is none. – Robert Israel Aug 10 '14 at 21:05
• Forgot I was dealing with logs where ln(a+b) can be contracted to ln(ab). – Cetshwayo Aug 10 '14 at 21:09
• @Cetshwayo That's incorrect as well. $\ln a + \ln b = \ln ab$. But $\ln(a+b) \neq \ln a + \ln b$. You cannot simply distribute the log of a sum. However, the sum of logs (to the same base) can be expressed as the log of a product. – Deepak Aug 11 '14 at 0:37

## You cannot distribute logarithms

You are confusing your identities. You are most likely confusing the identity $\ln(ab)=\ln a+\ln b$. In general $\ln(a+b)\neq\ln a+\ln b$. Also, note that if you are using a calculator, you do not need to simplify the expression; you can punch $\ln(1+\sqrt 3)$ straight into your calculator (but don't forget the brackets!).

• It seems something of a leap to conclude that the identity that the original poster was thinking of was $\ln(ab)=\ln a+\ln b$. More likely it was the false identity that says everything is linear. ${}\qquad{}$ – Michael Hardy Aug 10 '14 at 21:03
• Thank you! The post has been edited to sound less certain :P – Pauly B Aug 10 '14 at 21:15

As someone previously stated, logarithms are not linear. It can be confusing to keep the properties straight, but a good logic check is that if:

$$\log(a+b) = \log(a)+\log(b)\\ \rightarrow \log(a)=\log(0+a)=\log(0)+\log(a)\space\space(a\neq0)\\\rightarrow \log(a)~ DNE ~~ \forall a$$ Which is obviously incorrect.

Hint: Use the Taylor series of $\ln(1+x)$ at the point $x=1$. See related problem.

• That expansion is only valid for $|x|<1$. – preferred_anon Aug 10 '14 at 21:04
• @DanielLittlewood: You are right. I just noticed that. Corrected. – Mhenni Benghorbal Aug 10 '14 at 21:05
• @DanielLittlewood: It is corrected. – Mhenni Benghorbal Aug 10 '14 at 23:10
• I'm somewhat unhappy with the suggestion. You have to first calculate $\ln 2$ as the degree zero term of the Taylor series of $\ln(2+x)$ (which is what you are now suggesting). Then you need to substitute $x=\sqrt3-1$ to all the terms, so you also need to calculate $\sqrt3$ to a high precision. – Jyrki Lahtonen Aug 11 '14 at 5:57