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56

The first is $7^{91}\times 343$. The second is $7^{91}\times(9/7)^{91}$. Since $\frac{9^3}{7^3}\gt 2$, it follows that $(9/7)^{91}$ is much much bigger than $343$.


25

$$\array{808 &\gt& 9 \\ 101 \cdot 2^{3} &\gt& 3^{2} \\ 101^{3} \cdot 2^{3} &\gt& 101^{2} \cdot 3^{2} \\ \left((101\cdot 2)^{3}\right)^{101} &\gt& \left((101\cdot 3)^{2}\right)^{101} \\ 202^{303} &\gt& 303^{202}}$$


21

We have $a^b<b^a$ iff $b\log(a)<a\log(b)$ iff $\frac{\log(a)}a<\frac{\log(b)}b$. Consider the function: $$f:x\mapsto\frac{\log(x)}x:\mathbb R^+\to\mathbb R$$ Then: $$f'(x)=\frac{1-\log(x)}{x^2}$$ Hence $$f'(x)>0\quad\mathrm{iff}\quad x<e$$ Hence $f$ is decreasing for $x>e$ and this proves $f(303)<f(202)$, hence $303^{202}<202^{303}$.


21

$$7^{94} = 7^{10} 49 ^{42} < 7^{10} 54 ^{42} = 7^{10} 8^{14} 9^{63} < 9^{10} 9^{14} 9^{63} = 9^{87} < 9^{91} $$


15

$Log(9^{91})=91\cdot Log(9)=86.836068359$ $Log(7^{94})=94\cdot Log(7)=79.4392157613$. Hence $ 9^{91}$ is bigger.


11

The binomial theorem states that $$\sum_{k=0}^n \binom{n}{k} x^k=(1+x)^n$$ Putting $x=1$ gives $$\sum_{k=0}^n \binom nk=2^n\\ \sum_{k=1}^n\binom nk=2^n-1$$


8

André already nailed it, but here's another way. The following inequalities are equivalent:\begin{align}7^{94} &< (7+2)^{94-3} \\ 9^3&<(1+2/7)^{94} \\ 3\log3&<47\log(1+2/7),\end{align} and by the Maclaurin expansion of $\log(1+x)$, the latter follows from \begin{align}3\log3&<94\left(\frac17-\frac1{49}\right) \\ ...


8

The function $f(t)=t+2^t$ is strictly increasing in $\mathbb{R}$. So from $x+2^x=y+2^y$ it directly follows that $x=y$. And once you find that, it's easy to proceed.


8

Notice that for $f(x)=\sqrt x \ln x$ you have $$f'(x)=\frac{\ln x}{2\sqrt x}-\frac1{\sqrt x}.$$ Now by mean value theorem $$f(n+1)-f(n) = f'(\theta_n)$$ for some $\theta_n$ such that $n<\theta_n<n+1$ (by Mean Value Theorem). This means that for $n\to\infty$ we have $\theta_n\to\infty$. If we notice that $\lim\limits_{x\to\infty} f'(x) = 0$ we get that ...


7

Standard trick when things are difficult due to "mixed" terms like this: Write $$y_n = \sqrt{n + 1} \ln(n + 1) - \sqrt{n} \ln(n + 1) + \sqrt{n} \ln(n + 1) - \sqrt{n} \ln n$$ Now the first two terms can be combined as \begin{align*} \ln(n + 1) \big(\sqrt{n + 1} - \sqrt{n}\big) &= \ln(n + 1) \frac{1}{\sqrt{n + 1} + \sqrt{n}} \\ &\approx \frac{\ln ...


6

$9^{91} \div 7^{94} = (\frac97)^{94} \div 9^3 > (1+\frac27)^{7 \times 13} \div 3^6 > (1+2)^{13} \div 3^6 = 3^7$ which is way bigger than $1$.


6

I assume you are asking if there are other functions, $f$, such that $f'(x)=f(x)$. Well, we can write $1=f'(x)/f(x)=\log(f(x))'$. This implies that $\log(f(x))=x+c$, so that $f(x)=e^{x+c}=ke^x$ where $k=e^c$. Thus all functions are multiples of the exponential function. EDIT:: I am going to try to answer your second question. The functions $f(x)=ke^{ax}$ ...


6

Let's consider the meaning of $\sum_{k=1}^n \left(\begin{array}{c} n \\ k \end{array} \right).$ You are interested in the number of ways to construct a non-empty subset from the original set of $n$ elements. Consider the set with $n$ elements, for each element, we have two options, to include it in one of the subset or not. If we allow our subset to be ...


6

Taking all the rounds together (including the $0^{th}$), you have formed all combinations with any of the five letters taken or not, which you can do in $2\cdot2\cdot2\cdot2\cdot2$ ways. (Equivalently, all five bits numbers from $00000$ to $11111$, which is exactly $2^5$.) For perfect rigor, one should show that there are no repetitions nor omissions.


6

We have $$2^{3a}=3\implies 2^{3ab}=5\implies 2\times 2^{3ab}=10\implies 2^{3ab+1}=10$$ Thus $$10^c=2^{c(3ab+1)}$$ But we know that $2^{3ab}=5$ so we are asked to solve $$c(3ab+1)=3ab\implies c=\frac {3ab}{3ab+1}$$


5

It isn't different. There's a minus sign introduced by the derivative of (1-x), but there's also an odd power of (1-x), which is the second minus sign when compared to (x-1)


4

$ 24 \equiv 5 \bmod 19 $ $ 23 \equiv 4 \bmod 19 $ $ 5 \cdot 4 \equiv 1 \bmod 19 $ $ 5^{-31} 4^{32} \equiv 4^{31} 4^{32} \equiv 4^{63} \equiv 4^9 = 2^{18} \equiv 1 \bmod 19 $


4

We are comparing $202^{303}$ and $303^{202}$. $202^{303}$ is equal to $202^{202}$ * $202^{101}$. $303^{202}$ is equal to $(202 * 1.5)^{202}$ which is equal to $202^{202}$ * $1.5^{202}$ Now, we can divide out the $202^{202}$ from both sides which yields $202^{101}$ versus $1.5^{202}$. $1.5^{202}$ can be written as $2.25^{101}$ (squaring the inside, thus ...


4

Actually, on second thought this is a perfect problem for one to tinker, think about, and come up with a reason on one's own and to try to figure out why. So I sort of regret giving the answer. Before you read my answer try figuring things out on your own. Look at $2^12^2...... 2^{200}$ and try $2*4*8*16*32*64*128*.....*2^{200}$ and think about what ...


3

I voted for André's answer, but here's another approach, using a different bit of maths. Note that $7^{94} = 7^3 \times 7^{91}$. $9^{91} = (7 \times \frac{9}{7})^{91}$, where $\alpha = \frac{9}{7} = 1 + \frac{2}{7} > 1$. So $$ \frac{7^{94}}{9^{91}} = \frac{7^3}{\alpha^{91}}. $$ What do we make of $\frac{7^3}{\alpha^{91}}$? Well, $7^3 = 49 \times 7 = ...


3

$$\begin{align} \left(9\over7\right)^3={729\over343}\gt2 &\implies\left(9\over7\right)^{15}\gt2^5\gt7\\ &\implies9^{15}\gt7^{16}\\ &\implies9^{90}\gt7^{96}\\ &\implies9^{91}\gt7^{94} \end{align}$$ Another proof, using the general inequality $\ln(1-x)\lt-x$ for $0\lt x\lt1$ and the numerical inequality $7\lt2^3\lt e^3$: ...


3

With perhaps a little less arithmetic, $2^2=4\equiv23\pmod{19}$, and $4\times5=20\equiv1\pmod{19}$, so $24\equiv5\equiv4^{-1}\equiv2^{-2}\pmod{19}$. By Fermat's little theorem, $$23^{32}=2^{2\times32}=2^{64}\equiv2^{64-7\times18}\equiv2^{-62}\equiv2^{-2\times31}\equiv24^{31}\pmod{19}$$


3

After some calculation you have \begin{align*} 24 &\equiv 5 \bmod 19 \\ 24^2 &\equiv 25 \equiv 6 \bmod 19 \\ 24^4 &\equiv 36 \equiv -2 \bmod 19 \\ 24^8 & \equiv 4 \bmod 19 \\ 24^{16} & \equiv 16 \equiv -3 \bmod 19 \end{align*} Now multiply: $$24^{31} \equiv 5\cdot 6 \cdot (-2) \cdot 4 \cdot -3 \equiv (30) \cdot (24) \equiv 11 \cdot 5 ...


3

Note that $$ \begin{align} 1\le\frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}} &=\left(1+\frac1n\right)^{\sqrt{n}}(n+1)^{\sqrt{n+1}-\sqrt{n}}\\ &=\left(1+\frac1n\right)^{\large\frac{n}{\sqrt{n}}} (n+1)^{\large\frac1{\sqrt{n+1}+\sqrt{n}}}\\ &\le e^{\large\frac1{\sqrt{n}}} \left(\sqrt{n+1}^{\frac1{\sqrt{n+1}}}\right)^2 \end{align} $$ The Squeeze Theorem ...


3

We can do a bit more. Starting with what you wrote $$\ln (y_n)=\ln \frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}=\sqrt{n+1}\ln(n+1)-\sqrt{n}\ln(n)$$ rewrite $$\sqrt{n+1}\ln(n+1)=\sqrt n \sqrt{1+\frac 1n}\Big(\ln(n)+\ln(1+\frac 1n)\Big)$$ and now use Taylor series $$\sqrt{1+\frac 1n}=1+\frac{1}{2 n}-\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\ln(1+\frac ...


3

HINT: $$2^{2(3a-1)}=5^{2b-3}$$ Take logarithm to find $$2(3a-1)\log2=(2b-3)\log5$$


3

$3^{-4}=({3^4})^{-1}$ But $3^4=(3^2)^2=9^2=81$ So we get the answer is $81^{-1}=\frac{1}{81}$


2

We present an approach here that relies only on a standard inequality and the squeeze theorem. To that end, we proceed. First, note that we can write the limit of interest as $$\lim_{n\to \infty}\left(\frac{(n+1)^{\sqrt{n+1}}}{n^\sqrt{n}}\right)=\lim_{n\to \infty}e^{\sqrt{n+1}\log(n+1)-\sqrt{n}\log(n)} \tag 1$$ Next, we analyze the exponent on the ...


2

As you said the $420^{1337}$ contributes 1337 zeros and the $20160^{4646}$ contributes 4646 zeros so lets focus on the $900!$. In $900!$ we need to consider how many 2's and 5's there will be. Clearly there will be more 2's than 5's so the limiting factor for creating zeros at the end will be 5's. In $900!$ there will be $\frac{900}{5}=180$ numbers which ...


2

First notice that $3^9 = 19683 > 16807 = 7^5$ (this can be calculated manually). Thus $9^9 = (3^2)^9 = 3^{18} = (3^9)^2 > (7^5)^2 = 7^{10}$. It follows that $9^{91} > 9^{90} = (9^9)^{10} > (7^{10})^{10} = 7^{100} > 7^{94}$.



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