New answers tagged

1

Note that $$\frac d{dx}e^{ax}=ae^{ax}$$ We also have that, for $n\in\mathbb N$ $$\frac{d^n}{dx^n}e^{ax}=a^ne^{ax}$$ or, one could assume that this works for $n\in\mathbb R$, and put into notations, $$D^ne^{ax}=a^ne^{ax}$$ This is the result we want to get, probably through the methods you are given. I will say the above result should be correct via ...


0

Note that, if $f(x)$ is infinitely often differentiable and $f(0) \neq 0$ then $\frac{1}{f(x)}$ is infinitely often differentiable in $x=0$. This follows by induction, since you can show by induction that the $n$-th derivative of $\frac{1}{f(x)}$ is of the form $\frac{p(f, f^\prime, ..., f^{(n)})}{f^{2n}}$ with a polynomial $p$ of $n+1$ variables. Apply ...


5

Hint: Look at $g(x) = (e^x-1)/x$ instead. Then $g$ equals a power series that converges on all of $\mathbb R,$ hence is $C^\infty.$ Show that $g$ is never $0.$ Therefore …


0

We need to assume that both $\cos a, \sin a$ are positive in order to evaluate this limit. This means that $a$ is in first quadrant. Assuming that this is so, we can simply assume that $\cos a = A, \sin a = B$ where $A, B$ are fixed positive numbers with $A^{2} + B^{2} = 1$ we can put $x = 2 + h$ so that as $x \to 2$ we have $h \to 0$. Then we have ...


0

Just for clarification Simple Interest Simple interest is calculated by multiplying the loan amount (e.g. 1000 monetary units) by the interest rate (e.g. 5%) by the number of payment periods over the life of the loan (e.g. 24 months). The thing to keep in mind here is that the interest rate may be expressed in terms of an annual rate, e.g. 5% per annum, ...


0

That shouldn't give you a $\pm$ scenario, because the range of $x$ in a function $y=-2log_2(x)$ would be $(0,\infty)$. Therefore, even though you have $2^y=x^2$, you would want your x to be always positive, and that leads to a unique y value, so you would not actually deal with the $\pm$ scenario.


0

Given $y=-2\log_2(x)$, you have $\frac{y}{-2}=\log_2(x)$ so $2^{-(y/2)}=x$. Now you can choose values of $y$ like you did before and plug them in to get $x$-values. If you wanted to move the $-2$ into the log, you would get $y=\log_2(x^{-2})=\log_2\left(\frac{1}{x^2}\right)$, not $y=\log_2(x^2)$. Then $2^y=\frac{1}{x^2}$, so $x^2=\frac{1}{2^y}$. Now you're ...


11

The function $z \mapsto e^{-z}$ doesn't satisfy the conditions. Actually the limit $$\lim_{z\to \infty} e^{-z}$$ doesn't exist. (For exemple let $z$ go to infinity on $i\mathbb{R}$) Hence it is not a counterexample. The proposition is correct, by Liouville's theorem the function must by $1$ but then the limit is not $0$ so there are no such functions.


0

What about this? if x is positive, $$4^y=x^{-2} \rightarrow$$ $$y\ln 4=-2\ln x \rightarrow$$ $$\ln x =-y \ln 2\rightarrow$$ $$x =e^{-y \ln 2}$$ if x is negative, $$x =-e^{-y \ln 2}$$


0

Hint: You need to use the identity: $$a^{-b}=\frac1{a^b}$$ So then you have: $$4^y=\frac1{x^2}$$$$4^y*x^2=1$$$$x^2=\frac1{4^y}$$ so: $$x=\pm\frac1{\sqrt{4^y}}$$


0

Solving for $x$ gives two solutions $$ 4^y = x^{-2} \iff \\ 4^y = 1/x^2 \iff \\ x = \pm 1/\sqrt{4^y} $$


2

Note that $\cos kz = \mathrm{Re}(\cos kz + i \sin kz) = \mathrm{Re}(e^{ikz})$. Thus: $$ \sum_{k \geq 1} e^{-tk} \cos kz = \sum_{k \geq 1} \mathrm{Re}(e^{-tk})\mathrm{Re}(e^{ikz}) = \sum_{k \geq 1} \mathrm{Re}(e^{k(-t + iz)}) = \mathrm{Re}\left(\sum_{k \geq 1} \left(e^{-t + iz}\right)^k\right) $$ which is a geometric series.


2

$$\lim_{x\to \infty}\left(\frac{ax+1}{bx+2}\right)^x=\lim_{x\to \infty}\left(\frac ab\right)^x\left(\frac{1+\frac{1/a}{x}}{1+\frac{2/b}{x}}\right)^x \tag 1$$ Using the limit definition of the exponential function, we see that $$\lim_{x\to \infty}\left(\frac{1+\frac{1/a}{x}}{1+\frac{2/b}{x}}\right)^x=e^{1/a-2/b}$$ However, we have $$\lim_{x\to \infty} ...


1

Equations with messy powers are usually dealt with better by taking logs of everything in sight. So instead of looking for $\lim_{x \to \infty} f(x)$, look for $\lim_{x \to \infty} log(f(x))$ Once you know that, you can easily find the original limit easily. It is easy to start. $$\begin{align}\log(f(x)) &= \log((\frac{ax + 1}{bx + 2})^x) \\ &= x ...


1

We assume that $y$ is always positive. Let $z$ be the logarithm of $y$. Which base? It does not matter. I will use the natural logarithm $\ln$, but others might use the logarithm to the base $10$. Note that if $y=KC^x$ then $z=\ln y=\ln K+(\ln C)x$. Suppose that we are given values $x_1,x_2,\dots, x_n$ of $x$, and the associated values $y_1,y_2,\dots,y_n$ ...


1

Use: $$\log_a(b)=\frac{\ln(b)}{\ln(a)}$$ When $a$ and $b$ are positive: $$\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$$ So, we get when $a$ and $x$ are positive ($x\ne1$ and $a\ne1$): $$\log_a\left(x^a-x\right)-\log_a\left(\frac{x^a-x}{a}\right)=\frac{\ln\left(x^a-x\right)}{\ln(a)}-\frac{\ln\left(x^a-x\right)-\ln\left(a\right)}{\ln(a)}=$$ ...


2

Use: 1) $\log_a X - \log_a Y=\log_a \frac{X}Y$ 2) $\log_a a=1$ $$\log_a (x^{a}-x)-\log_a \Big(\dfrac{x^{a}-x}{a}\Big)=$$ $$=\log_a\frac{(x^{a}-x)}{\Big(\dfrac{x^{a}-x}{a}\Big)}=\log_a a=1$$


7

This solution uses the following facts: For every real number $t$, $\left(1+\frac{t}n\right)^n\to e^t$ when $n\to\infty$. For every real number $t$, $1+t\leqslant e^t$. For every real number $t\geqslant-\frac12$, $1+t\geqslant e^{t-t^2}$. One is asking to prove that the limit of ...


2

We will take Taylor series of $$f_n(x) = \sum_{i=1}^n \left( \frac{i+x}{n} \right)^n$$ We are aiming for $\frac{e^{x+1}}{e-1} = A (1+x+\frac{x^2}{2} + \dots)$ where $A = \frac{e}{e-1}$. The $x^0$ coefficient of the sum is $$f_n(0) = \sum_{i=1}^n \left(\frac{i}{n} \right)^n$$ If we differentiate the sum, we obtain $$f_n'(x) = \sum_{i=1}^n \left( ...


1

As already said in comments and answers, there is no analytical solution to this equation and numerical methods are required. Considering $$f(x)=x^{x-5}-5$$ the first derivative is given by $$f'(x)=x^{x-5}\left(1-\frac{5}{x}+\log (x) \right)$$ which cancels at $$x_*=\frac{5}{W(5 e)}\approx 2.57141$$ $W(z)$ being Lambert function. Using a calculator, you ...


1

Let $$f(x)=\ln(1-x)+x$$ We know, $f(1/6)<0$. Moreover, $$f'(x)=1-\frac1{1-x}<0$$ for all $0<x<1$. So, $f$ is decreasing. Thus, $f(x)<f(1/6)<0$ for all $x<1/6$, i.e. $l>3$.


5

You wrote $-1^0$, so the calculator interpreted it as $-(1^0)$ and not as $(-1)^0$. Use parentheses.


2

Problems of the form $x^x=a$ can be solved for $x$ using the Lambert W function. There is an example on the wikipedia. Problems of the form $x^{x+y}=a$ cannot be solved using the Lambert W function. If the Lambert W function cannot solve an exponential equation, don't expect any logarithms or anything nice if you are looking for a closed form solution. ...


4

Integrate both sides from $0$ to $x$ and add $1$.


0

Well, first off you need $x >0$, not just $ x \geq 0$. So, you know $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, which is, by definition, $\lim_\limits {k \to \infty} \sum_{n=0}^k\frac{x^n}{n!}$. Let $a_n = \frac{x^n}{n!}$, and $b_k = \sum_{n=0}^k\frac{x^n}{n!} = \sum_{n=0}^k a_n$. Clearly, $a_n > 0$, since $x>0$. Moreover, $b_{k+1} = b_k + a_{k+1} ...


0

I would use Taylor theorem with the Mean-value forms of the remainder, it is straightforward You can check https://en.wikipedia.org/wiki/Taylor%27s_theorem


1

In short: no, it is not. Using the Taylor expansions at 0 means that the terms you are neglecting (the $O(t^2)$) are only negligible in a neighborhood of $0$. In particular, when $t\to\infty$, they actually may — and will —dominate... Now, if you really want to use Taylor series at $0$, the usual trick is to make appear a parameter that tends to $0$ when ...


0

I've edited it at the user's notification. $$3^{x+4}-5(3^x)=684 \implies 3^x\cdot3^4-5(3^x)=684 \implies 3^x(3^4-5)=684 \implies (76)3^x=684 \implies 3^x = \frac{684}{76} \implies 3^x = 9$$ You can take it from here.


0

First, I urge you to graph ceiling function on a calculator to get a good grasp. x=1 is one counterexample. There are infinitely many. So, the exponent rule does not apply to the ceiling function.


-1

You factor out the $3^x:$ $$(1-5)(3^x)=680$$ $=$ $$-4(3^x)=680$$ Now, you can solve this equation with calculator.


0

I'm not sure if you mean $3^{x+4}-5(3^x)=684$ or $3^x+4-5(3^x)=684$. If you mean the first, then rewrite $3^{x+4}$ as $3^x\cdot3^4$ and then factor $3^x$ out. If you mean the second way, subtract $4$ from each side and then factor out $3^x$. Hope this helps.


2

There is a "closed-form" expression in terms of the Lambert W Function. We begin with the equation $$y(t)=e^{-t/\tau y(t)} \tag 1$$ Let $z=-t/\tau$ and let $W=\frac{z}{y(t)}$. Then, upon rearranging $(1)$ we find that $$z=We^W \tag 2$$ Noting that $(2)$ defines the Lambert W, we have immediately that $$\bbox[5px,border:2px solid ...


0

The car is $1-.08=.92$ of its initial value after every year. Thus, we can set up a function of the form $y=Ar^t$, where $y$ is the final value, $A$ is the initial value, $r$ is the rate of decay, and $t$ is the time parameter. Substituting our values, we find $$y=23000(.92)^t$$ To solve for the value of the car in $20$ years, set $t=20$.


1

Assuming the actual function is $$x(t) = 4e^{-0.2t}$$ We are asked to find the half-life time, i.e., the point in time $t$ where $$x(t) = \frac{x(0)}{2}$$ Plugin in above equation gives us $$4e^{-0.2t}= \frac{4e^{0}}{2}$$ $$4e^{-0.2t}= \frac{4}{2}$$ $$4e^{-0.2t}= 2$$ $$e^{-0.2t}= \frac{1}{2}$$ Now we take the $\ln$ logarithm of both sides, since on the ...


1

Your answers for the first two questions are correct. For part three you need to solve the equation $2=4e^{-0.2t}$. Try applying the log function to both sides.


5

The use of logs is actually unnecessary here, since by coincidence all bases in the equation can be rewritten as a power of a common base (namely, $2$). We can use exponent laws to simply both sides into the form $2^{X} = 2^Y$ and infer that $X = Y$. Indeed, observe that: \begin{align*} 16^{5a - 1} \cdot 256^{3a} &= 128 \\ (2^4)^{5a - 1} \cdot (2^8)^{3a} ...


0

$16^{5a−1} \times 256^{3a} =16^{5a-1}\times 16^{2(3a)}=16^{5a+6a-1}=16^{11a-1}=128$. Now, take a logarithm with base $16$. This yields $11a-1=\dfrac{7}{4}$. Now, you can solve this for $a$.


10

Let's take a log and see what happens: $$\log(x^a) = \log(x^b)\Longrightarrow a\log x = b\log x \Longrightarrow (b-a)\log x = 0.$$ There are two cases to consider here. I'll let you figure out what those are.


0

Hint: $$y(t)=y(t=0)e^{kt}$$, in which $t$ is time in days and $y$ is weight in kg. You have two parameters that need to be determined $y(t=0)$ (initial weight) and $k$ (growth rate). You also have to data points. $$y(t=7)=120=y(t=0)e^{7k}$$ $$y(t=30)=150=y(t=0)e^{30k}$$ Devide both equations to get the value for $k=1/23\ln(15/12)=1/23\ln(5/4)$ and plug ...


1

HINT Well, the answer is $x= \pm 1$. Realize that the function on the left is increasing for $x \ge 0$ and decreasing for $x \le 0$.


4

HINT: $$(5+\sqrt{24})(5-\sqrt{24})=1$$ Let $(5+\sqrt{24})^x=y\iff(5-\sqrt{24})^x=\dfrac1{(5+\sqrt{24})^x}=?$ Now solve for $y$ Now if $\displaystyle u^m=u^n,$ either $\displaystyle m-n=0,u\ne0; $ or $\displaystyle u=1$ or $\displaystyle u=-1,m-n$ is even


1

Let's assume the interest is compounded. After 1-year in the bank you have: $$1000+.07(1000)$$ $$=1.07(1000)$$ After another year we have, $$1.07(1000)+.07[1.07(1000)]$$ $$=1.07[1.07(1000)]$$ $$=(1.07)^2(1000)$$ See a pattern? Can you prove the pattern you see will always hold?


0

Here is another approach using differential equations: Let $Y$ solve $\dot{Y} = XY$ subject to $Y(0) = I$. Note that the unique solution is given by $Y(t) = e^{tX}$. Let $d(t) = \det Y(t)$, then $\dot{d}(t) = \det(Y(t)) \operatorname{tr} (Y^{-1}(t) X Y(t)) = \operatorname{tr} X \det Y(t) = \operatorname{tr} X d(t)$. Hence $d(1) = d(0) e^{\operatorname{tr} ...


4

If $X$ is similar to $Y$ with $X = P^{-1}YP$ then $$ \operatorname{Tr}(X) = \operatorname{Tr}(Y) \implies \exp(\operatorname{Tr}(X)) = \exp(\operatorname{Tr}(Y)), \\ \exp(X) = P^{-1}\exp(Y)P \implies \det(\exp(X)) = \det(\exp(Y)) $$ so we can assume that $X \in M_n(\mathbb{C})$ is in Jordan canonical form. In particular, $X$ is upper triangular with the ...


2

As said in comments, only numerical methods would be able to give the solutions. If the equations are $$(x-1)^2+(y+1)^2=36\tag 1$$ $$y=1+e^{-\frac{x^2}{10}}\tag 2$$ use $y$ as given by $(2)$ and plug it in $(1)$. You then need to search for the zeros of $$f(x)=(x-1)^2+\left(2+e^{-\frac{x^2}{10}}\right)^2-36$$ Using inspection or graphing, you will notice ...


0

This is not really an answer to your question but could provide some intuition. We have \begin{align*} e^x &= 1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots \end{align*} Thus \begin{align*} \frac{d}{dx}e^x &= ...


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 $f(x)=ke^{ax}$ ...


0

The amount you have at the beginning is $b$. After $1$ year, you have $a+b(1+c)$. After $2$ years, you have $a+(a+b(1+c))(1+c)$. After $3$ years, you have $a+(a+(a+b(1+c))(1+c))(1+c)$. etc. After $N$ years, you have $$b(1+c)^N+a(1+c)^{N-1}+a(1+c)^{N-2}+\cdots+a(1+c)+a$$ $$=\left(b+\frac{a}{c}\right)(1+c)^N-\frac{a}{c}$$ For a given $d$, in general you ...


0

You can't, not with elementary functions. As soon as your unknown is both outside and inside exp/log functions, it's hopeless - there's no closed form in terms of elementary functions. You have to attack the problem numerically (iteration, bisection, Newton's method,...). In this case, you can express the solution through the Lambert's W function, but that's ...



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