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46

Yes. They are the same thing. When exponents get really really complicated, mathematicians tend to start using $\exp(\mathrm{stuff})$ instead of $e^{\mathrm{stuff}}$. For example: $e^{x^5+2^x-7}$ is kind of hard to read. So instead one might write: $\exp(x^5+2^x-7)$. Note: For those who use Maple or other computer algebra systems, e^x is not usually the ...


29

Yes. The purpose for the notation $\exp$ is twofold: It allows one to talk about the exponentiation function itself, without specifying a particular input. For example, one can write that $\exp$ is a homomorphism from the additive group on $\mathbb{R}$ to the multiplicative group on $\mathbb{R}$. One may also say that $\exp$ and $\log$ are inverses. It ...


17

As other answers say, in your homework (and, indeed, in most places in mathematics) there is no difference. I have seen a beginning textbook first defining a certain function $\exp(x)$, then proving certain properties of it, and finally using those properties to motivate calling it $e^x$.


14

There is not. Consider as one example $$f(2^4, 2^4) = 2^{4\cdot 4} = 2^{16} =65536.$$ And you also want $$f(4^2, 4^2) = 4^{2\cdot 2} = 4^4 = 256.$$ But the left sides are equal (because $2^4=4^2$) while the right sides are not. So the arguments of the function do not have enough information to produce the results you want.


9

Using Bernoulli's Inequality, we get $$ \begin{align} \frac{(a+1)^{a+1}}{a^a}\frac{(a-1)^{a-1}}{a^a} &=\frac{a+1}{a-1}\left(1-\frac1{a^2}\right)^a\\ &\ge\frac{a+1}{a-1}\left(1-\frac1a\right)\\ &=\frac{a+1}a\\[8pt] &\gt1 \end{align} $$ Therefore, $$ \frac{a^a}{(a+1)^{a+1}}\lt\frac{(a-1)^{a-1}}{a^a} $$ which says that $\dfrac{a^a}{(a+1)^{a+1}}$ ...


8

You only need to recognize from exponent laws that $$x^{\frac 13}\times x^{\frac 12}=x^{\frac 13 + \frac 12}=x^{\frac 56}.$$ I think you are referring to the denominators of the fractions in the exponents. Yes, those denominators are different, but you can always find the common denominator first before adding the fractions: $$\frac 13 + \frac 12 = \frac 26 ...


7

I agree with these two answers, but I want to add one thing: Well defines. $e$ is some (positive) number, so (without knowing the function $\exp$), you can compute $e^n$ for $n \in \mathbb{N}$ – just multiply $e$ $n$ times with itself. You can also compute $e^{-n} = \frac{1}{e^n}$ and even $e^\frac{p}{q} = \sqrt[q] e^p$ (for $n, q \in \mathbb{N}, p \in ...


7

Hint. You may just observe that, for $x >0$, the function given by $$f(x)=\log \left(\dfrac{x^x}{(x+1)^{x+1}} \right) $$ is decreasing. Indeed, we have $$ f(x)=-x \log\left(1+\frac 1x\right)-\log (x+1) $$ and $$ f'(x)=- \log\left(1+\frac 1x\right)<0, \qquad x>0. $$


7

A commonly used set of algorithms is known as CORDIC: COordinate Rotation DIgital Computer. Basically, it uses a bunch of bitshifts, adds, subtracts and look up tables. Its good for cases where you don't have hardware multipliers and what not. You can also truncate series as well. An interesting thing is that you can often do some math and identify which ...


6

The reciprocal of $x$ is simply $\frac1x$. So the reciprocal of $(-\tfrac12)^k$ is $\frac1{(-\tfrac12)^k}$. You may also recall from the exponentiation laws that $(\frac1x)^y=x^{-y}$, and that the repciprocal of $x^y$ is $\frac1{x^y}=x^{-y}$. This allows us to rewrite $(-\tfrac 12)^k$ as $(-2)^{-k}$, and then the reciprocal of this is $(-2)^k$. If $k$ is ...


6

The basic rules you need to use are these: $$ \alpha a^n+\beta a^n+\cdots+\omega a^n=(\alpha+\beta+\cdots+\omega)\cdot a^n\tag{1} $$ and $$ a^n\cdot a^m=a^{n+m}.\tag{2} $$ Now let's consider what you have in the numerator: $$ 2^{20}+2^{20}+2^{20}+2^{21}=2^{20}+2^{20}+2^{20}+\underbrace{2\cdot 2^{20}}_{\text{by $(2)$}}=\underbrace{(1+1+1+2)\cdot ...


5

The function as initially described fails to exist by the definition of a function: for any given input (or input "vector") $x$, there must exist a single output $f(x)$. Here, the quickest counterexample for any such function $f(a^b,a^c)=a^{bc}$ is to note that $2^4=4^2$ and then any combination of these produces multiple outputs for a single specified ...


5

Use Fermat's Little Theorem. Since $13$ is a prime and $\gcd(3,13)=\gcd(5,13)=1$, we have, $$3^{13-1}=3^{12}\equiv 1\pmod{13}\quad\textrm{and}\quad 5^{13-1}=5^{12}\equiv 1\pmod{13}$$ $$\therefore\quad 3^{12}+5^{12}\equiv 1+1\equiv 2\pmod{13}$$


4

I would be surprised if they actually used Taylor series. For example, the basic 80x87 "exponential" instruction is F2XM1 which computes $2^x-1$ for $0 \le x \le 0.5$. I don't think the implementation is documented, but if I were programming this, I might use a minimax polynomial or rational approximation: the following polynomial of degree $9$ ...


4

When we want to take a number to an integer power, we can simply multiply repeatedly and/or take reciprocals. When we want to take a positive real number to an arbitrary power, we usually use logarithms. Logarithms and Exponentials When considering raising a positive real number, $a$, to an arbitrary power, $x$, we usually compute as follows $$ ...


4

Although I agree with the answers already provided that in this situation (and indeed in most other ones in mathematics) there is no difference between the two notations, I would like to add the following for completeness: In manifold theory (most particularly Lie Group theory or Riemannian geometry), the exponential map $\exp$ is a map from a tangent space ...


4

$$\dfrac{(2\ln a- \ln b - 5\ln c)}{2}$$ $$ =\dfrac{(\ln a^2- \ln b - \ln c^5)}{2} $$ $$= \dfrac{\left(\ln \dfrac{a^2}{b} - \ln c^5\right)}{2} $$ $$= \dfrac{1}{2}\ln \dfrac{a^2}{bc^5} $$ $$= \ln \dfrac{a}{\sqrt{bc^5}}$$


4

This isn't a definitive answer (unfortunately I haven't been able to find any history on the subject when I've looked it up before), but I once noticed that you can write something like $2^{x^2+1}$ and it is clear in which order one must compute the operations. But if you divide that quantity by two it becomes $2^{x^2+1-1}=2^{x^2}$. Why should one suddenly ...


3

$(a^b){}^{{}^c}$ is the same as $(a^b)^c=a^{bc}$. If it was evaluated left-to-right, we'd have $a^{b^c}=a^{bc}$, so power towers would do nothing. The way it's usually done, from right to left, has $a^{b^c}\ne a^{bc}$. Ultimately, that's what it comes down to. EDIT: As mentioned elsewhere, right-to-left is the only way for something like $e^{x^2+1}$ to ...


3

As noted by Musafa you have: $$ \dfrac{5^{2n-1}-5^{2n-2}}{5^{3n-3}-5^{3n-2}} $$ Now note that $5^{2n-1}=5\times 5^{2n-2}$ and the same for $5^{3n-2}=5 \times 5^{3n-3}$. Using distributivity you can simplifies the fraction and finally you find the result of Wolfram. ( if you have some problem I can help). $$ ...


3

The key, I think, is to appeal to the definition of the absolute value function: $|x| = x$ if $x > 0$, and $|x| = -x$ if $x \le 0$. Your desired implication then follows immediately.


3

It's always true that $|x|\geq0$. So if $-x=|x|$, then this means $-x\geq0$. That is equivalent to $x\leq0$.


3

While both expressions are generally the same, $\exp(x)$ is well-defined for a really large slurry of argument domains via its series: $x$ can be complex, imaginary, or even quadratic matrices. The basic operation of exponentiation implicated by writing $e^x$ tends to have ickier definitions, like having to think about branches when writing $e^{1\over2}$ or ...


3

What you want to prove is: given an irrational number $b$, then one of the following numbers : $$r^{\frac{1}{b}}\ \ \ \ \ \ r\in \Bbb Q $$ is irrational, this seems to be an attainable result for what we know nowadays about the transcendence of numbers, we have for example the following theorem: Six Exponentials Theorem:Let $(x_1,x_2)$ and ...


3

It's, unfortunately, not a particularly well-defined problem, as infinite power towers aren't always well defined. However, if we want to apply algebraic techniques anyhow, notice that we can write it as $$2x^{\left(2x^{2x^{2x\ldots}}\right)}=4$$ where the inner expression on the left is equal to four for a solution, giving $$2x^4=4$$ which is easier to ...


3

We have an associative operation here – it is the function composition $\circ$. For any functions $f : X \to Y$ and $g : Y \to Z$ we have $g \circ f : X \to Z$, defined by $(g \circ f)(x) := g(f(x))$ for all $x \in X$. I heard some authors defined it the other way around, $(f \circ g)(x) := g(f(x))$. I think my version is more common.) This operation is ...


2

The ultimate answer can only be given by the maker of the calculator, but there are several strategies: The value can be calculated through the series expansion. The value can be interpolated from a stored table. The value can be reduced to the value of a smaller/larger argument by basic operations; for example $e^{2x} = (e^x)^2 = e^x\times e^x$. In ...


2

$$ 2x^{2x^{2x^{..}}}=4 \longrightarrow 2x^4=4 $$ $$ x^4=2 $$ I think you can do the rest.


2

Without FLT and, perhaps, more basically: observe that $$3^3=1\pmod{13}\;,\;\;5^2=-1\pmod{13}$$ thus $$3^{12}+5^{12}=\left(3^3\right)^4+\left(5^2\right)^6=1^4+(-1)^6\pmod{13}=1+1=2\pmod{13}$$


2

hint: $$5\cdot 2^{n-1} = 5\cdot 2\cdot 2^{n-2} = 10\cdot 2^{n-2}, 5\cdot 3^{n-1} = 5\cdot 3\cdot 3^{n-2} = 15\cdot 3^{n-2}$$, and combine like terms.



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