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So $1000(2^n+3^n)\le 4^n+5^n$. I take $1000\cdot2^n\le 4^n$ and $1000\cdot3^n\le5^n$ so that adding both gives the inequality, theorem in the ordered fields. So $1000\cdot2^n\le2^2n$. This leads me to an $A$ for $n = 10$ as $1000\cdot2^n < 2^n\cdot2^n$ so $1000\leq2^n$ for $n=10$. But this does not count for $1000\cdot3^n<5^n$. I think to see that $1000\cdot3^n<3^n\left(\frac{5}{3}\right)^n$ so $1000<\left(\frac{5}{3}\right)^n$. so $n = \log_{\frac{5}{3}} (1000)$. This leads me to an $A = 14$ which is my final answer.

Any other suggestions.

A similar problem is $\displaystyle\frac{(1.01)^n}{n^{147}} < 100$. This is tougher. Any hints

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    I mentioned it elsewhere. This exercise, and the second mentioned, appear in the handbook, where log, binomials and so on have not been introduced yet. So, it demands for a simple, perhaps although rudimentary solution.2012-07-09

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For the similar problem, the inequality is false. If we take the base $10$ log of each side, we get $n \log 1.01 - 147 \log n \lt 2$ or $ .00432n - 147 \log n \lt 2$ Now we just need to take $n$ large enough. $n=10^6$ is easily enough.

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    @Ignace: Generally there is no algebraic solution without Lambert's W function. You are usually into some form of numeric root finding, discussed in any numerical analysis text.2012-05-16