Are exponentials with jointly prime bases linearly independent?

Question

Do the exponentials $3^x$, $4^x$ and $7^x$ have the property that for no $\alpha, \beta, \gamma \neq 0$:

$\alpha 3^x + \beta 4^x = \gamma 7^x$

What if the bases are not jointly prime, like if they were 4 and 8?

Answer 1

Yes, those expononential functions are linearly indipendent.

• If one of $a, b, c $ is zero we have clearly that the other two coefficients are zero.
• If two of $a, b, c$ are zero same thing.

So we suppose $abc \ne 0$

We can prove that $\exists x: c7^x > a3^x+b4^x$

In fact $c7^x > (a+b)4^x $ (this implies $c7^x > a3^x+b4^x$) iff $x>\frac{ln \frac{a+b}{c}}{ln(7/4)}\space \square$

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One can generalize. Let $a_1, ..., a_n$ be real number such that $a_i>1$ and $a_1>a_2>...>a_n$ (strictly increasing!). We can easily show that $\exists x: \lambda_1a_1^x>\lambda_2a_2^x+...\lambda_na_n^x$ ($\lambda_i$ are arbitrary positive constants).

In fact $\exists x$ such that (why?) $$\lambda_1a_1^x\stackrel{(*)}{>}\left(\sum_{i=2}^{n}\lambda_i \right)a_2^x> \sum_{i=2}^{n}\lambda_ia_i^x $$ You could prove (*) which is very easy. The other inequality comes from hypothesis.

In particular this answer to your second question.