Mathematics Stack Exchange
Mathematics Stack Exchange

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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Exponential growth with growth factor <2

Is there an $a$ with $1
Christopher Rose
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Subtracting exponents with same base

If $2^4 - 2^3 = 2^3$ and $2^5 - 2^4 = 2^4$, then is below a rule of subtracting exponents with similar base and exponents which are just $1$ away from each other? $$A^e - A^{e-1} = A^{e-1}$$ I will also like to get an visual intuition of why this…
LoveWithMaths
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Operations Equivalence

Subtracting a number relates to summing a number, as dividing by a number relates to multiplying by a number, as ??? relates to powering by a number. May I put it in a more mathematical manner: $x-y \leftrightarrow x+y\space;…
Leonardo Bohac
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Solving $a^{a^{a^{\cdots\infty}}}=2$?

I was thinking about the problem $a^{a^{a^{\cdots}}}=2$, I thought of taking $x = a^{a^{a^\cdots}}$, then we get $a^x = x$ but we know as per the question that $x=2$ so we have to solve for $a^x=x$ or $a^2=2$ which we get $a=\sqrt{2}$, but this is…
BAYMAX
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How to Solve the Exponential Equations where unknown appears in both base and exponential part?

Recently I encountered an exponential equation in form of $a^x + x^a = b^x$, where both $a,b$ are integers. The tricky part is $x$ appears in both the base and the exponential part. Is this kind of equation solvable? For instance, $6^x + x^6 =…
hzh
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Variable Powers

I'm getting a lot of these type of questions, and it's getting increasingly frustrating to solve them by trial and error. I've tried logarithms and derivatives, and they either aren't working or I'm applying them wrong. $$2^x(4-x)=2x+4$$ The only…
DynamoBlaze
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Product of exponents property

$a, b, c, d, e, f$ are natural numbers. $a$ and $b$ are co-prime(relatively prime). If this equation is true: $$ a ^ c \times b ^ d = a ^ e \times b ^ f$$ Then does this mean that both these equations are always true: $$c = e ,d = f$$ ? Is it…
BorticFan
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Figure out all positive integers n with consecutive + integers a,b,c.

When $2018^n$ = $a^4$ + $b^4$ + $({b^2+c^2})^2$, then what is the possible positive integers n be?
허인행
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How to find $y^{(y^2-6)}$?

$$\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = y$$ $$x≠2$$ $$y^{(y^2-6)} = ?$$What is $y^{(y^2-6)}$? Could you please explain to me how to solve this question step by step?
chsdwn
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List in the function exponent meaning

I get a weird exponential inside the function that appears as the result of an integral in Mathematica. Namely, the function $Hypergeometric1F1^{(1,0,0)}[1, 1, -x^2]$ has the 3-component list in the exponential. I have searched everywhere to find…
Mihailo
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Comparing two exponentiations

$b^n$ where the base $b$ is a positive integer greater than $1$ and the exponent $n$ is a rational number in simplified form. How would one compare (resulting in <, =, or >) two such exponentiations without evaluating the exponentiatoins, and…
Arlen
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$9^x=5$, solve for $5\cdot27^{-x-1}$

Question: Given $9^x=5$ Compute the value of $5\cdot27^{-x-1}$ My attempt: $9^x=(3^x)^2=5\Rightarrow3^x=\sqrt{5}$ $5\cdot27^{-x-1}=3^{2x}\cdot(3^{3-x}\div3^{3})=3^{2x}\cdot3^{-x}=3^{x}=\sqrt{5}$ My answer is incorrect and I wonder why.
student
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Fourth term in the expansion of $(1-2x)^{3/2}$

Calculate the fourth term in the expansion of $(1-2x)^{3/2}$ I first tried to use binomial theorem , but of course fractions in combinations cannot be calculated (or is it that I don't know how to). And then is we add $r$ from $0$ to $
Sri
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How can $1 + 2^{1/x} = 0$ be solved?

I'm trying to solve an equation: $ 1 + 2^\frac{1}{x} = 0 $ My first thought was to solve this by with the help of logarithms $\log_2 -1 = \frac{1}{x}$, but when I wrote it down, I saw a meaningless gibberish. Then I tried to solve this, by…
d.k
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Rules on exponents

What's wrong with this ? $$-2=(-2)^{1}=(-2)^{\frac22}=(-2)^{2\times\frac12}=((-2)^2)^{\frac12} = 4^{\frac12} = 2$$ Note that Wolfram Alpha gives : $-2=(-2)^{2/2} = (-2)^{2\times\frac12}$, and on the other hand $((-2)^2)^{\frac12}=2$. So, based on…
Sephi
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