# Tagged Questions

Questions about exponentiation, the operation of raising a base $b$ to an exponent $a$ to give $b^a$.

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### negative fraction exponent and division

Quick question on how to handle negative fraction exponents when differentiating: I have this problem to differentiate. $$x^{2/3} + y^{2/3} = 1$$ So my textbook and I both did the first thing the ...
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### Why does $\frac{1}{4}x^{-3/4} = \frac{1}{4x^{3/4}} = \frac{1}{4\sqrt[4]{x^3}}$?

This is taken from Khan Academy, I don't understand how these equate: $$\frac{1}{4}x^{-3/4} = \frac{1}{4x^{3/4}} = \frac{1}{4\sqrt[4]{x^3}}$$ How come the minus was remove from the original exponent?...
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### Name for inequality about sums of exponents with same base

Is there a name for the following inequality regarding sums of exponents which share a base? $$\text{For all integers b \geq 2, n \geq 1,} \\ \sum_{i=0}^{n-1}{b^i} < b^n$$
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### Is it possible to transform different interest rates (applied over different times) into one interest rate?

I have a fixed income asset that has the following payment schedule. Is it possible to transform the multiple interest rates being applied to different periods into one interest rate and one period? I ...
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### Enough differences of powers of natural numbers equals a constant

Let $n_k$ be a sequence. We create a new sequence by taking the difference of consecutive terms in $n_k$. So the terms of the new sequence $a_k$ is defined as $a_i=n_{i+1}-n_i$. This is the difference ...
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### How to differentiate x^(1/x)?

How to differentiate the following? $$x^{\frac{1}{x}}$$ (I know the answer is $\frac{1-\ln(x)}{x^{2-\frac{1}{x}}}$, but I do not understand how to get there) Attempt at solution I believe the ...
I always thought that $(a^b)^c=a^{bc}$... I am confused about why the order of exponents seems to matter in this particular case: $((x-0.5)^2)^{1/3}$ is not the same as $(x-0.5)^{2/3}$ ...
Show that inequality holds for every n, k. $$\ { n \choose k} \leq \frac{n^k}{k!}. ( \, 1-\frac {k}{2n} ) \,^{k-1}$$ I used Stirling's formula but I stucked, which is \ { n \choose k} = \frac{n^...