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Questions tagged [exponentiation]

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

Exponentiation is a mathematical operation which produces a power $a^n$ from a base $a$ and an exponent $n$. The objects involved are usually numbers, but the procedure can be generalized to matrices, elements in algebraic structures, sets, etc.

4326 questions
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How to prove $(a \cdot b)^n = a^n \cdot b^n$?

Assuming $a, b \in \{\mathbb{R^-,R^+}\}$ and $n \in \mathbb{R}$. We often see the statement: $$(a \cdot b)^n = a^n \cdot b^n$$ Why we get the statement? How to prove this? There should pay attention, the $n \in \mathbb{R}$, not $\mathbb{N}$.
244boy
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$(q^{2^{n+1}})^2$ question to understand another question

I found question, that is primary question for my problem. Can't ask my question via comment to the second answer, because have not enough reputation. In proving of $$(1+q)(1+q^2)(1+q^4)\dots(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$$ I got till…
Pavel Stepanov
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Why rational exponents expressed in the simplest form are defined as roots but those not in the simplest form are not?

If $b$ is a nonzero real number and $p/q$ is a positive rational number, then, if $p/q$ is expressed in lowest terms, $b^{p/q}=\sqrt[q]{b^p}$. Why if $p/q$ is not expressed in lowest terms, $b^{p/q}$ may or may not equal $\sqrt[q]{b^p}$?
Muhammad
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what is the solution to this equation: $1 + 3^{x/2} = 2^x$?

what is the solution to this equation: $1 + 3^{x/2} = 2^x$ ? The answer is $x = 2$. I want to know the process.
Mosiur Rahman
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What is the difference between $\dfrac{1}{3^{-2}}$ and $3^{-2}$

Stupid question but how in the world does 1/3^-2 not be the same thing as 3^-2? The first answer I got is 9, the second one I got is 0.111... Isn't the negative power just 1/3/3? What difference does it make when I do 1/1/3/3?
Solomon Mao
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What happens when a is negative in $ a^{\frac{m}{n}} $.

What happens when a is negative in $ a^{\frac{m}{n}} $. This question may seem silly but I got confused by it. If you take the example $2^{\frac{4}{3}}$ Method 1: $\sqrt[3]{2^4} \approx 2.51984$ Method 2: ${(\sqrt[3]{2})^4} \approx 2.51984$ But if…
Aditya Gupta
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Sum of n with factoral exponent

I'm not sure what this is called, but the application is in counting the number of nodes in a consistently branching structure. For example, $5$ nodes branch into $5$ nodes each, each branching again, etc, $5$ times overall. I think the math would…
Karric
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Reversing formulas containing exponents

I have these 4 formulas e = 4 * n^0.6 e = n^3 e = 1.2 * n^3 - 15 * n^2 + 100*n - 140 e = 5 * n^0.75 They take in n and result in e How on earth would I reverse this to take in e and result in n. I'm not very math savy is another problem. I mean I…
June
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Why is $0.5^x + 0.5^x = 0.5^{x-1} $

I believe that the following is always true: $0.5^x + 0.5^x = 0.5^{x-1} $ but I do not know why. I've tried to prove it but am unsure how. Is it always true? And is there a proof?
Rina
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multiplication with exponentials

How is this correct? $$\hbar\omega…
Lopey Tall
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How to write $12 (N / m)^2$ in long form

I want to write the value $12 (N/m)^2$ in long form like "12 Newtons per meter squared". However, I believe that because of the order of operations, my long form value will be interpreted as $12 N/(m^2)$. How should I write this value in long form?…
johnnyodonnell
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$\;x^{52}=x^{-4}\cdot r^7\;$ what will be the value of r?

If $x^{52}=x^{-4}\cdot r^7$ then what will be the value of $r$?
Mad Dawg
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Solution to equation $2^x $ = $x^8$

I was able to solve this equation using graphical methods, but cannot figure out a mathematical solution to the equation. What approach should be taken to solve it?
Sid
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Simplifying $\left\{\left[(2/9)^4\times(3/14)^4\right]^4:(-1/7)^2\right\}\times\left[(-5/6)^3:(5/18)^3\right]^3$

$$\left\{\left[\left(\frac29\right)^4\times\left(\frac3{14}\right)^4\right]^4:\left[\left(-\frac17\right)^2\right]\right\}\times\left[\left(-\frac56\right)^3:\left(\frac5{18}\right)^3\right]^3$$ I've been trying to simplify this expression for $3$…
user602325
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Solve y(a^y-1)=b

Trying to solve this equation for $y$ leads me into horrible thickets of logarithms that also seem unsolvable. $$y(a^y-1)=b$$ $a$ and $b$ are constants. Is there a simple solution of the form: $$y = f( a, b, c... )$$ where c... represents possible…
Iron Pillow
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