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14

Elucidate the problem by using the substitution $u = 2^x$, then you have $$u + \frac{1}{u} = 2$$ Multiply throughout by $u \neq 0$ to get $$u^2 +1 = 2u \iff u^2 - 2u + 1 = 0$$ This is an easy quadratic to solve, you should get $u = 1$ and hence you need only solve $2^x = 1 \iff x = 0$.


11

Hint: By AM-GM inequality , $2^x+2^{-x} \geq 2 \times \sqrt{2^x \times 2 ^{-x}} = 2 $ When $2 ^x = 2^{-x}$ the equality holds. Alternative solution: Easy to show that $x = 0$, is one possible solution. $$\frac{d(2^x+2^{-x})}{dx} = ( 2^x -2^{-x}) \log 2$$ which positive over $(0,+\infty)$, negative over $(-\infty,0)$


8

Assuming $x$ to be natural, $$ 5^x-3^x=(5-3)(5^{x-1}+3\cdot5^{x-2}+3^2\cdot5^{x-3}+\cdots+3^{x-1})=4\cdot2^x\\ \implies (5^{x-1}+3\cdot5^{x-2}+3^2\cdot5^{x-3}+\cdots+3^{x-1})=4\cdot2^{x-1} $$ Each term in LHS looks like $3^{n}5^{x-1-n}>2^n2^{x-1-n}=2^{x-1}$. There are $x$ number of terms in LHS. If there are $4$ or more terms in LHS, each of them would ...


7

This result is commonly shown via Taylor series, as explained in the comments, and is well-known. I'd like to offer a different sort of proof, for those who are interested, that I believe is easier yet less well-known. Consider the second order linear differential equation $$y''=-y$$ We know the most general solution is: $$y = A\cos{x}+B\sin{x}$$ But $$y = ...


6

Hint: Divide both sides by $5^x$, and notice that the left side is strictly decreasing, since it is the sum of two strictly decreasing functions, and the right side is constant. It follows that our equation has at most one solution. Can you find it ? ;-$)$


6

For the first problem you will need a little bit of Chinese Remainder Theorem. You want to find the remainder of the stacked exponential modulo $10^5 = 2^5 \times 5^5$. Consider the two prime divisors separately. As $\phi(2^5) = 16$ we have that if $r_1$ is the remainder of $5^{5^{5^{5}}}$ modulo $\phi(2^5) = 16$ then $5^{5^{5^{5^{5}}}} \equiv 5^{r_1} \pmod ...


6

$$2^{340}\cdot 5^{340} \lt \dfrac{5^{496}}{1985}$$ is equivalent to $$2^{340}\lt \frac{5^{496-340}}{5\times 397}=\frac{5^{155}}{397}$$ which is equivalent to $$397\cdot 2^{340}\lt 5^{155}$$ Since $397\lt 1024=2^{10}$, it is sufficient to prove that $$2^{350}\lt 5^{155}$$ which is equivalent to $$2^{70}\lt 5\cdot 5^{30}$$ which is equivalent to $$5\gt \left(...


6

A contrived solution: Write $$\frac{e^{x\ln2}+e^{-x\ln2}}2=1=\cosh(x\ln2),$$ then $$x\ln2=\cosh^{-1}1=0.$$ Another one: By inspection, $x=0$ is a solution. The derivative of the LHS is $$2^x\ln2-2^{-x}\ln2,$$ which is positive for $x>0$ and negative for $x<0$, so the function is monotonic on both sides and there are no other roots. Yet ...


5

hint: Take log we have: $(2x^2-3x+1)\log |x| \leq 0$. You can have two cases here to continue, but simple.


5

Observe that $$-(32^\frac15)=-(2^{5\cdot\frac15})=-2$$ And $$(-32)^\frac15=(-2)^{5\cdot\frac15}=-2$$ I hope your doubt is resolved now.


5

$$-32^{\frac{1}{5}}=(-32)^{\frac{1}{5}}=(-2)^{{5}^{\frac{1}{5}}}=(-2)^{5\cdot \frac{1}{5}}=-2$$


5

Intuitively, the way to think about exponentiation is by considering exponential growth/ decay. Let's say we have a bacterial sample growing on an agar medium. And we happen to know that this bacteria will triple in population size every minute. Then, if $P_0$ is the initial population, the population at any later time will be $$P(t) = P_0\cdot 3^t$$ ...


4

When $y> 0$, you have $$\sqrt{\frac x{y^2}}=\frac{\sqrt{x}}{\sqrt{y^2}}=\frac{\sqrt{x}}{y}.$$ When $y<0$, you have $$\sqrt{\frac x{y^2}}=\frac{\sqrt{x}}{\sqrt{y^2}}=\frac{\sqrt{x}}{-y}.$$


4

There is no systematic way of solving such transcendental equations. In this particular case, we can get better insight by performing a transformation to get rid of the double exponentiation. $$4=x^{x^4}=(x^4)^{x^4/4}=t^{t/4}.$$ Then $$t^t=4^4$$ and an obvious solution is $t=4$ corresponding to $x=\pm\sqrt2$. By the study of the function $t^t$, we can ...


4

Let $J$ be the $n\times n$ Jordan block with eigenvalue $\lambda$. I'll assume we're working over $\mathbb C$ (or at least in characteristic $0$). Claim: If $\lambda\neq 0$ then the Jordan normal form of $J^m$ is an $n\times n$ Jordan block with eigenvalue $\lambda^m$. If $\lambda=0$ then the Jordan normal form of $J^m$ is $r$ blocks of size $q+1$ and $m-r$ ...


4

Multiplication is proven to be associative; that means you can calculate any part of a series of it first (= setting and removing parenthesis wherever you want), and the result is always the same. All the different variants you noted are correct, and they are all identical.


3

Exponents less than 1 are better understood in terms of fractions rather than decimals: $3^4=3^{2+2}=3^2•3^2$ Therefore: $3^1=3^{\frac{1}{2}+\frac{1}{2}}=3^\frac{1}{2}•3^\frac{1}{2}=(3^\frac{1}{2})^2$ What value, when multiplied by itself, is equal to 3? The only possibility is ±$\sqrt{3}$. This is why we understand an exponent of $\frac{1}{2}$ to be ...


3

I will interpret the question to mean: For function $f_{a,b}(x) = a^x + b^x$, do parameters $\alpha$ and $\beta$ exist such that $f_{a,b}(x) =\alpha \beta^x$? The answer is no. Assume that such parameters do exist. Let $x = 0$, to get $\alpha = 2$ and then let $x = 1$ to get $\beta = \frac{a+b}2$. Now, take any other $x$ to derive a contradiction. ...


3

Walking through the first problem, we effectively need to find $a \equiv 5^{5^{5^{5^{5}}}} \bmod 10^5$. This splits easily into finding $a_1 \equiv a \bmod 2^5$ and $a_2 \equiv a \bmod 5^5$ which can then be re-united with the Chinese remainder theorem. The order of $5 \bmod 32$ divides $\phi(32)=16$ (and actually we could say it divides $\lambda(32) = 8$ ...


3

I will assume that $a \ge 1$ and show that the only solution to $2^{2a+1}+2^a+1 = n^2$ is $a=4, n=23$. This is very non-elegant but I think that it is correct. I just kept charging forward, hoping that the cases would terminate. Fortunately, it seems that they have. If $2^{2a+1}+2^a+1 = n^2$, then $2^{2a+1}+2^a = n^2-1$ or $2^a(2^{a+1}+1) = (n+1)(n-1)$. ...


3

$r^{-1} = \frac{1}{r}$ That is why. This is because an exponent is the number of times the same number is multiplied together. So: $$x^4=x\times x\times x\times x$$ If we then divide by $x$, this is the same as multiplying by $x^{-1}$ Because $$\frac{x\times x\times x\times x}{x} = x^4\times x^{-1} = x^{4-1} = x^3$$


2

Let's write $n=x-y$. If $720!=3^x-3^y=3^y(3^n-1)$, then $3^n-1$ is divisible by the primes $17$, $31$, $43$, and $79$ (among others, of course). As it happens, $3$ is a primitive root for those primes. This means that $n$ is divisible by $16$, $30$, $42$, and $78$, hence by the lcm of these, or $$16\cdot15\cdot7\cdot39=65{,}520$$ But this is vastly ...


2

Hint: Analyse this function and try to prove $x=2$ is only solution $$f\left( x \right) =4\cdot 2^{ x }+3^{ x }-5^{ x }$$


2

$$\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$$ Is "the solution" to $ax^2 + bx + c$, provided that $a,b,c$ are real numbers and $a \ne 0$. $4a^2 \ge 0$ for all real numbers $a$. Since we are excluding $a = 0$, then $4a^2 > 0$. We can then say $$\sqrt{4a^2} = \begin{cases} 2a, & \text{if $a \ge 0$} \\ -2a, & \text{if $a < 0$} \...


2

You tried to use $2+3=5$ to no avail. A slightly different "flash of insight" approach would be to invoke Pythagoras: $3^2+4^2=5^2$.


2

Some tricks which are useful for modular exponentiation The intention of this post is to collect various tricks which can sometimes simplify computations of this type. (Especially when done by hand and not using computer or calculator.) This post is community-wiki, so feel free to edit it if you have some ideas for improvements. Using complement: $(c-a) \...


2

Officially, you should parse $1/4a^{-2}$ as $(1/4) \cdot a^{-2}$. The exponential is evaluated first, then the multiplies and divides from left to right. Certainly the $4$ belongs in the denominator. The question is whether the $a^{-2}$ belongs in the denominator. Given that the exponent is negative, I would think it likely that the author meant that. ...


2

You can try it by induction: For the base case is trivial that $1,2\in \mathbb{Z}$. Then suppose that for $k$you have that $a_{k},b_{k}\in \mathbb{Z}$. Finally, to prove the last step, check that $$(1+2i)(a_{k}+ib_k)=(a_k-2b_k)+i(b_k+2a_k)$$ So $a_{k+1}=a_k-2b_k$ and $b_{k+1}=b_k+2a_k$ where both are integers.


2

This could be called the binomial expansion or the binomial theorem. When $x$ is a positive integer, we have the well-known relationship $$(a+b)^x = \sum_{k=0}^x \begin{pmatrix} x \\ k \end{pmatrix} a^k b^{x-k}.$$ The quantity $\begin{pmatrix} x \\ k\end{pmatrix} = \frac{x!}{k!(x-k)!}$ is known as the binomial coefficient. When $x$ is non-integral, it ...


2

Dr. Graubner's answer is spot on but for a different approach that might be easier to use when you don't have a calculator (or are fairly bad with logs) you can take the following very rough approach. $10^{80}<16^{80}=2^{4*80}=2^{320}$ so $$\frac{2^{64^{64}}}{10^{80}}>\frac{2^{64^{64}}}{2^{320}}=2^{64^{64}-320}$$ Now $64^{64}-320$ is so close to $64^...



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