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Hot answers tagged exponentiation

89

$-x^2$, in every mathematical context I have seen, always means $-(x^2)$. So $-3^2 = -9$. On the other hand, when you plug in a value to an expression you don't just plug it in directly, you add parentheses first. For example, in the expression $7x$,if you plug in $x + 3$ for $x$, you get $7(x + 3) = 7x + 21$, not $7x + 3$. Similarly, plugging in $x = -3$ ...

54

One method is to set $x=y^2$ and rearrange this as a quadratic equation $$x^2-153x+1296=0$$

29

Using Bernoulli's inequality: $$(1+x)^n\ge 1+nx \qquad \forall \ \text{x\ge -1 and n\in\mathbb N_0 }$$ we have $$\left(1-\frac1{99}\right)^{49}\ge 1-\frac{49}{99}=\frac{50}{99}>\frac{50}{100}=\frac12$$ Therefore $$\frac{98^{99}}{99^{98}}=98\cdot \left(1-\frac1{99}\right)^{49}\cdot \left(1-\frac1{99}\right)^{49}>98\cdot ... 25 Here's the solution:$$y^4-153y^2 +1296 = 0 y^4 -144y^2-9y^2+1296 = 0 y^2(y^2-144) -9(y^2-144) = 0 (y^2-9)(y^2-144)=0 (y^2-3^2)(y^2-12^2)=0$$note that a^2 - b^2 = (a-b)(a+b) Can you see how now? I trust you can finish the rest 21 -3^2 is always -9. There's no ambiguity. And the odd/even rule is also true! x^n is always nonnegative when n is even, and x^n is the same sign as x when n is odd (when x is real). There's no contradiction, because "-3^2" isn't actually of the form x^n. See, x^n, when you substitute x=-3 and n=2, gives you "(-3)^2," not "-3^2." ... 14 6005 already explained why they don't contradict themselves. As an alternative answer to the first part of the question, "higher order" operators usually take precedence: exponentiation is applied before multiplication, which again is applied before subtraction. Whether you interpret unary minus in -x to be 0-x or (0-1)∙x it then follows that -x^2 ... 14 The inequality 99^{98} < 98^{99} is equivalent to$$\bigl( 1 + \frac{1}{98} \bigr)^{98} < 98 $$I'll prove by induction that$$\bigl( 1 + \frac{1}{n} \bigr)^n < n \quad\text{if $n \ge 3$}\quad\text{(it's false if $n=1$ or $2$).} $$The basis step n=3 is \frac{64}{27} < 3. So let's assume that$$\bigl( 1 + \frac{1}{n} \bigr)^n < n ...

12

The only "elementary" way I can think of is to write $99^{98} = (98 + 1)^{98}$ and then expand using the binomial expansion formula, and then show you get a sum of $99$ terms where each term is less than or equal to $98^{98}$, and the sum of the last two terms $98 + 1$ is strictly less than $98^{98}$. Then your sum is strictly less than $98 \cdot 98^{98} = ... 10 It's not a matter of syntax, it's a matter of operator precedence. In the absence of parentheses, exponentiation is executed first, then negation. Try this in Wolfram Alpha: -3^2. Then try (-3)^2. We have long agreed on these rules so that computers deliver consistent results on calculations involving various different arithmetic operations. As for the ... 9 No. Take$p = 73$as a counterexample.$4$has order$9 \pmod{73}$and we get that$4^n + 1$is periodic$\bmod 73$with cycle $$5, 17, 65, 38, 3, 9, 33, 56, 2.$$ In general, your claim fails if$\textrm{ord}_p(4)$is odd. 7 The accepted syntax is one that goes by the standard rules of operator precedence and associativity that most mathematicians, scientists and computer programmers have followed for decades if not centuries. Then, given $$-3^2 = -9$$ the squaring is done first, giving us$9$, and the negation is done second, resulting in$-9. If instead you have $$(-3)^2 = ... 7 While the convert-to-quadratic approach works here, and other answers have provided this approach, it's worth noting that rational solutions to quartics with integer coefficients work similarly to quadratics. That is, if you have a quartic of the form$$ ax^4+bx^3+cx^2+dx+e=0 $$where all coefficients are integers, then all rational solutions must be of the ... 7 Let$$\displaystyle f(x) = x^{\frac{1}{x}}\;,$$where x>0 Now$$\displaystyle f'(x) = x^{\frac{1}{x}}\cdot \left[\frac{1-\ln (x)}{x^2}\right]$$So here f'(x)>0 for x<e and f'(x)<0 for x>e\approx 2.714 So function f(x) is an Strictly Decreasing function for x>e\approx 2.714 So$$f(98)>f(99)\Rightarrow ... 7 I believe the simplest way to simplify your expression is with fractional exponents, though there are other ways. \begin{align} \sqrt[10]{32a^5} &= \left(32a^5\right)^{1/10} \\[2 ex] &= \left(2^5a^5\right)^{1/10} \\[2 ex] &= 2^{5\cdot 1/10}a^{5\cdot 1/10} \\[2 ex] &= 2^{1/2}a^{1/2} \\[2 ex] &= (2a)^{1/2} \\[2 ex] &= \sqrt{2a} ... 6\frac{98^{99}}{99^{98}}=\frac{98}{\left(1+\frac{1}{98}\right)^{98}}>\frac{98}{97}>1$$5 Hint: plot the graph of y=2^x and y=4x and shows that the only other solution is between 0 and 1. 5 I assume your friend and you both know that the expression (-1)^x only makes sense if x is an integer. That said, you have$$(-1)^{(-n)} = \frac{1}{(-1)^n} = \left(\frac{1}{-1}\right)^n = (-1)^nI believe all steps only use equalities you learnt in school: For all a, b we have a^{-b} = \frac{1}{a^b} If \frac{c}{d} is a fraction, then ... 5 Let u = a \, x^{2} then du = 2 a \, x \, dx. In the integral presented the differential unit is part of the integrand. With this in mind then \begin{align} I &= \int_{0}^{\infty} x \, e^{-a \, x^{2}} \, dx = \frac{1}{2a} \, \int_{0}^{\infty} e^{- u} \, du = \frac{1}{2a}. \end{align} Now, if integration by parts is considered then the following is ... 5 One typical way to write integration by parts is to identify your original integral as \int u \, dv and then we can use the identity\int u \, dv = uv - \int v \, du$$Where we form du from u by differentiating u with respect to the independent variable, and we form v from dv by antidifferentiating. In your case, if I am taking you rightly you ... 4 It appears that you mean to ask "2^{987}-2^{986}=?". Note that 2^{987}=2\cdot 2^{986} and that 2x-x = x. So:$$2^{987}-2^{986} = 2\cdot 2^{986}-2^{986} = (2-1)2^{986} = 1\cdot 2^{986}=2^{986}$$4 Let 4=x^{(x^2)}=\left((x^2)^{(x^2)}\right)^{x^2/2}$$x^2=y\implies y^{(y^2)}=16$$One of the value of y is 2$$x^{(x^2)}+x^{(x^8)}=x^2+x^{16}=2+(2)^8$$4 Since we are dealing with x^{1/n}, I assume that x\gt0. In that case, \lim\limits_{n\to\infty}x^{1/n}=1, and thus the series$$ \sum_{n=1}^\infty x^{1/n} $$diverges by the Term Test. 4 We could try and generalize a little bit and ask ourselves if n ^ {n+1} is always bigger than (n+1) ^ n where n is a natural number. This turns into verifying if the inequality n^{n+1} > (n+1)^n holds. The left hand side can be rewritten as n \cdot n^n and the inequality becomes n \cdot n^n > (n+1) ^ n Being n^n always positive, the ... 4 If one wishes to use logarithms to facilitate then we proceed accordingly. We have that$$x^{x^3}=3\implies x^3\log x=\log 3\implies x^3\log x^3=\log 3^3 \implies (x^3)^{x^3}=3^3$$Of course this result could be obtained instantly by simply cubing both sides of the original equation. Therefore a solution is evidently x^3=3 or x=3^{1/3}. Given ... 4 First observe that the last two digits of powers of 3 repeat with a cycle of 20 elements:$$1,3,9,27,81,43,29,87,61,83,49,47,41,23,69,7,21,63,89,67,\ldots$$Then it's sufficient to find the last two digits of 7^{2016} to arrive at the answer. The last two digits of powers of 7 repeat with a cycle of 4:$$1,7,49,43,\ldots$$Therefore$$7^{2016} ... 4 The best answer I can give is that there is no accepted syntax because it creates sufficient ambiguity to cause problems. Thus the rule should be: only use-3^2$if it is completely unambiguous what is meant due to context, otherwise use$(-3)^2$or$-(3^2)$to provide readers with unambiguous resolution. I have seen, in typesetting, the use of a smaller ... 4 In the context of an algebra class I believe an algebraic proof will suffice: $$-1a=-a$$this property is given before exponents are introduced. Now making the substitution$a=b^n$gives the algebraic result needed $$-1b^n=-b^n$$ For your example$b=3$and$n=2$gives$-1•3^2=-3^2$The left hand side being$-1•9=-9$The right hand side must be ... 4$-x^n$always means$-(x^n)$. We don't need a parentheses-free syntax for$(-x)^n$, especially when$n$is a literal number like 2, 3, etc., since$(-x)^n = (-1)^nx^n$, which is always either$x^n$or$-x^n$(depending on whether$n$is odd or even) (assuming$n$is a non-negative integer). So$(-x)^n$can always be re-written as either$x^n$or$-x^n$, ... 3 So the Bible isn't the only book that can be completely misunderstood when passages are taken out of context. Nor is it the only book to contain contradictions. At least if you find a genuine contradiction in an algebra textbook you won't be accused of being a devil worshiper. However, you have not spotted a genuine contradiction here. I'm not bringing up ... 3 I hope at least we can use the Binomial Theorem with integer exponents. I will also assume knowledge of the usual$\binom mn$notation for the coefficients, but that is not really necessary; you can rewrite the argument to do without that notation. Observe that for$m \geq 1$,$\$\binom{98}{m + 1} = \frac{98 \cdot 97 \cdot 96 \cdot \cdots \cdot (98 - m)} ...

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