Hot answers tagged

7

Exponents multiply in the sense that $$(a^b)^c = a^{bc}.$$ They do not multiply in the sense that $$a^{(b^c)} \ne a^{bc}.$$


6

Note that if $x\to 0^-$ and $t=-1/x$ then $x=-1/t$ and $t\to+\infty$, hence $$\lim_{x\to 0^-}\frac{e^{1/x}}{x^2} = \lim_{t\to\infty}\frac{e^{-t}}{(-1/t)^2} = \lim_{t\to\infty}\frac{t^2}{e^t}.$$ And the last limit should be a proud member of the collection of limits you know.


6

For all values of $x$, $e^x>0$ is true. This means that $a\cdot e^x > 0$ is true if and only if $a>0$.


5

Divide by $e^x$ $$(x-2)e^x < 0$$ $$\iff x - 2 < 0$$ This is valid since $e^x > 0$ $\forall x \in \mathbb{R}$


4

Note that $\ln(a^x)=x\ln(a)$ Thus $$ln(e^{2x}) = ln(4/3)$$ $$(2x)ln(e) = ln(4/3)$$ $$2x = ln(4/3)$$ since $\ln(e)=1$ $$x=\frac{1}{2}\ln(\frac{4}{3})$$


3

Note that the ceiling changes when its argument passes through an integer. This happens when the uniform variable passes through $1/n$ for some $n$. The ceiling will be $n$ when the uniform variable is between $1/n$ and $1/(n-1)$. The length of this interval is $\frac{1}{n-1}-\frac{1}{n}=\frac{1}{n(n-1)}$, and the probability of a uniform random variable on ...


3

Recall that for $a > 0$ $$ab < ac \iff b < c$$ Note that $e^x > 0 \ \forall x \in \mathbb R$


2

$$2^x+5^x=3^x+4^x \iff5^x-4^x=3^x-2^x$$ Lagrange's theorem applies (intermediate value theorem) functions: $$f:[4, 5]\to R,\ g:[2, 3]\to R,\ f(u)= u^x,\ g(v)=v^x$$ Exist $c\in [4, 5]$ and exist $ d\in [2, 3] $ so $5^x-4^x = xc^{x-1}$ and $3^x-2^x=xd^{x-1}$. The equation is written as equivalent: $$xc^{x-1}= xd^{x-1}.$$ It follows that equation solutions ...


2

Notice, a few things: $$\log_a(x)=\frac{\ln(x)}{\ln(a)}$$ $$\ln(e)=\log_e(e)=\frac{\ln(e)}{\ln(e)}=1$$ $$\ln(x)=\log_e(x)=\frac{\ln(x)}{\ln(e)}=\frac{\ln(x)}{1}=\ln(x)$$ $$\ln(a^x)=x\ln(a)\space\space\space\text{when}\space a,x\space\text{are positive}$$ $$\ln\left(\frac{a}{x}\right)=\ln(a)-\ln(x)\space\space\space\text{when}\space a,x\space\text{are ...


2

The answer should be $6.25$. \begin{align} & 4 \sinh (2 \ln 2) - \cosh(\ln2 ) \\ =& 2 \left(e^{2\ln2}-e^{-2\ln2} \right) - \frac{e^{\ln2}+e^{-\ln2}}{2}\\ =& 2(4-0.25)-\frac{2+0.5}{2}\\ =& 7.5-1.25 = 6.25. \end{align}


2

Hint Since $x \mapsto \log x$ is a strictly increasing function, applying it to both sides of the inequality gives that it is equivalent to $$x \geq e \log x .$$ There are at least two options here: (1) Rearrange the inequality as $\frac{\log x}{x} \leq \frac{1}{e}$, and analyze the expression on the l.h.s. of this equality. In particular, show that its ...


2

If you allow, you can solve for $x$ through Lagrange Inversion Theorem. $$y=x^{x+1}$$ Invert it... I now have time to work the problem out. $$f(x)=x^{x+1}$$ $$f(1)=1,f'(1)\ne0$$ $$f^{-1}(x)=1+\sum_{n=1}^{\infty}\lim_{w\to1}\frac{(x-1)^n}{n!}\frac{d^{n-1}}{dw^{n-1}}\left(\frac{w-1}{w^{w+1}-1}\right)^n$$ This is not reducible as far as I know, so you ...


2

Without using derivation, one can use the inequality $\ln(x)\leq x-1$ and its equivalent form that: $$ \ln(x)\geq 1-\frac{1}x. (\star) $$ Now suppose that there is $m$ such that $\forall x$ $\ln(x)\leq m(x-1)$. Note that this is equivalent to your problem. Indeed $m=\ln(a)$ and we can safely assume that $m>0$ (put $x=1$ and then $a>2$). So we have: $$ ...


1

If you set $t=x+1$, the inequality becomes $a^{t-1}\ge t$ and it's sufficient to analyze it for $t>0$, so it is the same as $$ (t-1)\log a\ge\log t $$ Consider the function $$ f(t)=(t-1)\log a-\log t $$ We want to see whether or not the function is nonnegative. If $0<a\le1$ it's obvious it isn't, since in this case $$ \lim_{t\to\infty}f(t)=-\infty $$ ...


1

I don't understand what you did, I (and Wolfram Alpha) would do this: $${d\over dx}\left(ye^y\right) = {d\over dx}\left(x\right)$$ $$\iff {d\over dx}\left(ye^y\right) = 1$$ $$\iff y'e^{y} + y'e^{y}y = 1$$ $$\iff y'\left(e^{y} + e^{y}y\right) = 1$$ $$\iff y'(x) = {1 \over e^{y} + e^{y}y}$$ $$\iff y'(x) = {1 \over e^{y}(1 + y)}$$


1

In that step when you first took the derivative of $y$ you should get $\frac{dy}{dx}= \frac{y}{(y+1)x}$, right?


1

You're correct: $$\begin{align} 4\sinh{2\log{2}} - \cosh{\log{2}} &= \cosh{(\log{2})} \left( 8\sinh{(\log{2})}-1 \right) \\ &= \frac{1}{2}(2+1/2)(4 \cdot 2 - 4 \cdot 1/2-1) \\ &= \frac{5}{4}(8-2-1) \\ &= \frac{25}{4} \end{align}$$


1

The order in which you do the operations is vital here. Unlike addition and multiplication, exponentiation is not associative: $$a^{(b^{\large c})} \neq \left(a^b\right)^c$$ for "most" values of $a$, $b$, and $c$. It is true that $2^{1/n} = 2^{n^{\large -1}}$, but that's because $2^{n^{\large -1}}$ means $2^{(n^{\large -1})}$. As you saw for yourself, ...


1

In general, let $X_1,\dotsc, X_n$ be iid eponential distribution with mean $1/\lambda$. Then the distribution of the minimum $M$ is $$P(M\leq m) = 1-P(M>m) = 1-(e^{-\lambda m})^{n} = 1-e^{-\lambda nm}.$$ Notice that this shows that $M$ follows an exponential distribution with mean $\frac{1}{\lambda n}$. In our case, $\lambda = 1/22.5, n =4$, and so $$E[M] ...


1

$$\ln\left(A\cdot B^X\right)=\ln\left(C\cdot D^X\right)$$ $$\Leftrightarrow\ln\left(A\right)+\ln\left(B^X\right)=\ln\left(C\right)+\ln\left(D^X\right)$$ $$\Leftrightarrow\ln\left(A\right)+X\ln\left(B\right)=\ln\left(C\right)+X\ln\left(D\right)$$ $$\Leftrightarrow X\ln\left(B\right)-X\ln\left(D\right)=\ln\left(C\right)-\ln\left(A\right)$$ ...


1

Let us suppose wlog $a,b,c,d>0$ (it seems obvious from what you said). Suppose moreover $b\neq d$. $$ ab^x=cd^x\Longleftrightarrow \frac ac=\frac{d^x}{b^x}=\left(\frac db\right)^x $$ Now let us take the logarithm in base $d/b$ which is positive and different from $1$: you get $$ \log_{\frac db}\left(\frac ac\right)=\log_{\frac db}\left(\frac ...


1

As mfl commented, a linear fit $y=a+bx$ looks more appropriate than $y=ae^{bx}$. For the linear model, as already given by mfl, we should get $$y=76.6228+0.0944818 x\qquad SSQ=7363.8\qquad R^2=0.995485$$ Because of the curvature for low values of $x$, a better fit could be obtained using $y=a+bx^c$. For such a case $$y=37.6705+0.57678 x^{0.804092}\qquad ...


1

Yes your solution is correct $$2^{-k} = \frac 1 n \implies n = 2^k \implies \log_{2} n = k$$


1

From the second line, $x= 3e^{2y} + 4$ $\Rightarrow 3e^{2y} = x-4$ $\Rightarrow 6e^{2y} = 2x-8$.


1

$3e^{2y} + 4 = x \implies 3e^{2y} = x - 4 \implies 6e^{2y} = 2x - 8 $ Therefore : $\frac{1}{2x - 8 }=\frac{1}{6e^{2y}}$


1

If you actually want to use a substitution to check your answer, then if $y=f^{-1}(x)$ and $f(x)=3e^{2x}+4$ then $$y=\dfrac12\log_e\left(\dfrac{x-4}{3}\right)$$ which will give $$\frac{1}{6e^{2y}} = \frac{1}{6e^{\log_e\left(\frac{x-4}{3}\right)}}= \frac{1}{6\left(\frac{x-4}{3}\right)}= \frac{1}{2x-8}$$ but this is not strictly necessary here



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