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I have seen the proof that $\{\ln(2),\pi\}$ is linearly independent

Found here:

Are $\pi$ and $\ln(2)$ linearly independent over rational numbers?

However, when trying to show that $\{\ln(2), \pi, 1\}$ is linearly independent (or dependent) I am having some trouble.

Here was my attempt: Suppose that $\{\ln(2), \pi, 1\}$ is linearly dependent then For $a,b,c$ be elements of $Q$ such that $a,b,c$ are non-zero.

$$a + b\ln(2) + c\pi = 0$$

$$e^{a+c\pi}(2^b) = 1$$

Once I got to this point I am feeling shaky about the $e^{a+c\pi}$. I realize that this is very close to Gelfond's Constant $(e^\pi)$, but I have tried to do some gymnastics with the algebra and I can't quite get it to a form where I recognize that this is transcendental (hence irrational). A point in the right direction would be highly appreciated!

Michael Hardy
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Jon Staggs
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    This follows from Baker's theorem. See my answer to your previous question about $\pi/4 \pm \ln 2$. – Sungjin Kim May 24 '17 at 23:05
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    @i707107 : Neither stating Baker's theorem nor linking to anything about it seems rather odd in this context. Here's a link: https://en.wikipedia.org/wiki/Baker%27s_theorem $\qquad$ – Michael Hardy May 24 '17 at 23:23
  • @MichaelHardy Thank you for the link. I wanted to remind him that I mentioned Baker's theorem in the answer to a former question. – Sungjin Kim May 24 '17 at 23:31
  • More interestingly, can we show that no (rational) linear combination of pi and ln(2) is rational? – Jacob Wakem May 24 '17 at 23:31
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    Not sure of this being complete, but integrals of the form $\int_0^1 \frac{x^m(1-x)^n}{1+x^2}dx$ yield results of the form $a+b\log(2)+c\pi$ That integrand is small but not zero, thus making $a+b\log(2)+c\pi=0$ impossible. – Jaume Oliver Lafont May 24 '17 at 23:32
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    Any nontrivial linear combination of $\pi$ and $\ln 2$ over algebraic numbers is transcendental. i.e. $a\pi + b\ln 2$ is transcendental for any algebraic $a$, $b$, not all zero. – Sungjin Kim May 24 '17 at 23:34
  • @i707107 you have proven to be my hero time and time again. Thank you! – Jon Staggs May 24 '17 at 23:39

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