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How can I solve $(1+\frac {0.05}{x})^x$ < $ \frac 1{100000}$?

I took natural log on both sides, but could not find anything..

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If $x>0$, you are raising a number greater than $1$ to a positive power, so your answer will be greater than $1$.

If $x=0$, the expression is undefined.

If $x<0$, you are raising a number less than $1$ to a negative power, so your answer will be greater than $1$.

Therefore, there are no (real-valued) solutions to your equation.

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What is your goal here? You have an inequality that can't be simplified a whole lot. Your inequality is clearly equal to $$100000(1+\frac{1}{20x})^x < 1$$

Expanding this we get $$20^{-x} (-20 x-1)^x (-x)^{-x}<\frac{1}{100000}$$ Or equivalently $$2^{-2 x-5} 5^{-x-5} \bigg(\Big(\frac{20 x+1}{x}\Big)^x-2^{2 x-5} 5^{x-5}\bigg)<0$$

and finally, if $x>0$ we have $$100000(20+\frac{1}{x})^x < 20^x$$

Source

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