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2

We have $$(8^x)^2+8^x-20=0$$ $$\implies8^x=\frac{-1\pm\sqrt{1-4(-20)}}2=-5,4$$ For real $x,8^x>0$ Then $8^x=4\implies x\ln8=\ln4$ But $\ln8=\ln(2^3)=3\ln2$ and $\ln4=\ln(2^2)=2\ln2$


6

Hint: Make the substitution $u=8^x$. Now the equation becomes $u^2+u-20=0$.


0

This is essentially another way of saying what sanjab has already said, but in a way that gives it a bit more intellectual context. Define a function as follows. $$\otimes : \mathbb{R}_{>0} \times \mathbb{R}_{>0} \longrightarrow \mathbb{R}_{>0}$$ $$p \otimes q = \exp(\log(p)\log(q))$$ Then $\otimes$ is associative and commutative. Its unit is $e$, ...


3

take the $\log$ for both sides to get $$\log (x^{log(a)})=\log (a^{\log(x)})$$ $$\log (a){log(x)}=\log( a){\log(x)}$$ It is clear that the $a$,$x$ should be positive


1

Yes: the general definition of $A^B$ is $\exp(B\log(A))$.


12

$$\large x^{\log(a)} = (e^{\log(x)})^{\log(a)} = e^{\log(x)\log(a)} = a^{\log(x)}$$ This only works if $x$ and $a$ are both positive real numbers.


2

Step 1: Compute the norm $$r:=\sqrt{4^2+16^2\cdot 5}$$ Step 2: Compute the tangent of the argument (and find the angle) $$\tan(\theta)=\frac{16\sqrt{5}}{-4}$$ Mind the sign of the real part $-4$ to get the appropriate angle. Step 3: Write the number $$re^{i\theta}.$$


1

Take $\nu=1-\lambda$. Since $\nu\in(0,1)$ we have $\nu=e^{-r}$ with $r>0$ and we have to prove that for $$f(x) = 1-e^{-rx} $$ we have: $$ \frac{f(x)-f(x-1)}{x-(x-1)} \leq \frac{f(x)-f(0)}{x-0}. $$ This follows from the fact that $f(x)$ is a concave function, since: $$ f''(x) = -r^2 e^{-rx} < 0.$$


2

The Problem The main problem when computing $e^{-20}$ is that the terms of the series grow to $\frac{20^{20}}{20!}\approx43099804$ before getting smaller. Then the sum must cancel to be $\approx2.0611536\times10^{-9}$. In a floating point environment, this means that $16$ digits of accuracy in the sum are being thrown away due to the precision of the large ...


0

We can always find the exponential of a $2 \times 2$ matrix ( with real or complex entries) without calculating eigenvalues or Jordan form. 1) Note that we can always put a $2 \times 2$ matrix in the form $A=hI+A'$ where: $h=\mbox{tr}A/2$, $\mbox{tr}A'=0$, $\mbox{det}A'=\mbox{det}A-\left(\dfrac{\mbox{tr}A}{2}\right)^2$; and $[hI,A']=0$ so that ...


0

The function $$f: \mathbb R \longrightarrow \mathbb R, f(x):= ax^b\cdot\exp(x)$$ extends to a holomorphic function on the complex plane $$F: \mathbb C \longrightarrow \mathbb C, F(z):= az^b\cdot\exp(z).$$ Global holomorphic functions are named entire functions. An entire function which is not a polynomial is named a transcendental function. As a ...


0

1) The regular double precision floating point arithmetic of Matlab is not sufficient to precisely calculate partial sums of this power series. To overcome that limitation, you can use the exact symbolic computation capabilities of the Symbolic Math Toolbox. This code x = sym(-20); i = sym(1 : 100); expx = sum(x .^ (i - 1) ./ factorial(i - 1)) sums the ...


2

Consider $f(x) = xe^x-e$. You have that $x = 1$ is a root. And: $$f'(x) = e^x + xe^x > 0,$$ for $x > 0$, so we don't have any positive roots other than $1$. If $x \leq 0$, you would have $e = \text{something negative}$, which can't happen.


6

The function $f(x) = xe^x$ does not have an inverse that can be expressed in terms of any finite algebraic combination of the usual functions. The two branches that form its inverse are known as the Lambert W function(s). The equation $$ y = xe^x $$ can be solved for $y$ using a numerical approach, if an approximate solution is desired.


2

The approach you mention is difficult, but possible. I have presented it in my blog post. The main steps are as follows: 1) Define $a^{b}$ (without using any logs or $e$) rigorously for $a > 0$ and any real $b$. 2) Show that $\lim_{a \to 0}\dfrac{x^{a} - 1}{a} = f(x)$ exists for all $x > 0$ and hence defines a function of $x$. This function is ...


0

It's easier to start from the first equation and prove that it's the same as the second one. From $$L := \lim _{a \to 0} \frac{e^a-1}{a}$$ set $e^a - 1 = t$, so that $a = \ln(1 + t)$: $$L = \lim_{t \to 0} \frac{t}{\ln(1 + t)} = \lim_{t \to 0} \left(\frac{\ln(1 + t)}{t}\right)^{-1} = \lim_{t \to 0} \left(\ln\left[\left(1 + t\right)^{1}\right]\right)^{-1} = ...


0

$$ f' = x f \iff \int \frac{ df}{f} = \int x dx \implies \ln |f| = x^2 + C \implies f(x) = \tilde C e^{x^2} $$ with $C \in \mathbb{R}$


1

$$(d/dx) \ln f(x) = \frac{f'(x)}{f(x)} = x.$$ Therefore $\ln f(x) = \frac{1}{2}x^2 + C$, so $f(x) = e^{\frac{1}{2}x^2 + C} = De^{\frac{1}{2}x^2}$, where $D$ is an arbitrary positive constant.


6

Note that by the Mean Value Theorem, we have $$ \frac{\tanh(x)}{x}=\mathrm{sech}^2(\xi)\tag{1} $$ for some $\xi$ between $0$ and $x$. Therefore, for $x\ne0$, $$ 0\lt\frac{\tanh(x)}{x}\lt1\tag{2} $$ Thus, for $x\ne0$, $$ \begin{align} x\frac{\mathrm{d}}{\mathrm{d}x}\left(e^{x-x^2/2}+e^{-x-x^2/2}\right) &=(x-x^2)e^{x-x^2/2}-(x+x^2)e^{-x-x^2/2}\\ ...


5

First, since both sides are even functions, it is sufficient to prove that $e^x+e^{-x} \leq 2e^{x^2/2}$ for all $x \geq 0$. Then we can change the problem to the equivalent problem of proving that for $x \geq 0$ $$ u(x) \equiv \frac{1}{e^x+e^{-x}} \geq \frac{1}{2}e^{-x^2/2} \equiv v(x)$$ We start by observing that $u(0) = v(0).$ Now in the immortal words of ...


19

$y = 2 e^{x^2/2}$ satisfies the d.e. $y' = x y$ with $y(0) = 2$, while $z = e^{x} + e^{-x} = 2 \cosh(x)$ satisfies $z' = \tanh(x) z$ with $z(0)=2$. Since $x \ge \tanh(x) $ for $x \ge 0$, Gronwall's inequality does the rest.


14

As @flawr points out, the L.H.S. is $\displaystyle \cosh x = \frac{e^x+e^{-x}}{2}$ has an infinite product representation: $$\cosh x = \prod\limits_{n=1}^{\infty} \left(1+\frac{4x^2}{\pi^2(2n-1)^2}\right) \le \exp \sum\limits_{n=1}^{\infty}\frac{4x^2}{\pi^2(2n-1)^2} = e^{x^2/2}$$


2

As said in comments and answers, there is no analytical solutions fo such an equation and numerical methods should be used. Probably, the simplest root-finding method is Newton which, starting from a reasonable guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)} {f'(x_n)}$$ This could be applied to the orginal equation ...


1

You might wanna try this: Let $7^x$ be $X$ and $3^x$ be $Y$. $\implies X+Y=XY$ Adding 1 to both sides, $\implies 1=(XY-X-Y+1)=(X-1)(Y-1)$ $\implies (7^x-1)(3^x-1)=1$ Perhaps here graphing is the best option. Even a rough graph could give the idea that one solution here is $- \infty$. Also, seeing the slope of the rough graph one can also think of the ...


1

Hint: Divide both sides by $21^x$, and use the fact that we now have a sum of monotonous functions being equal to something constant. Obviously, this can only happen for a single value of x, so the equation has only one solution. Which is what you were probably tasked to determine in the first place, i.e., the number of solutions, not their actual value, ...


4

You won't get a closed-form solution, but numerical methods can be used.


0

I can let $y=\frac{100}{1+2^{-x}}$. $$ 1+2^{-x} = \frac{100}{y}$$ $$ 2^{-x} = \frac{100}{y}-1$$ $$ 2^{-x} = \frac{100-y}{y}$$ $$2^x = \frac{y}{100-y}$$ $$x = \log_2 \left ( \frac{y}{100-y}\right)$$


0

$$g(x) = \frac{100}{1+2^{-x}}$$ $$\frac{100}{g(x)} = 1+2^{-x}$$ $$\frac{100}{g(x)} -1 = 2^{-x}$$ $$\ln\left(\frac{100}{g(x)} -1\right) / \ln(2) = -x$$ $$-\ln\left(\frac{100}{g(x)} -1\right) / \ln(2) = x$$


0

$g(x)=\frac{100}{1+2^{-x}}\implies$ $g(x)\cdot(1+2^{-x})=100\implies$ $1+2^{-x}=\frac{100}{g(x)}\implies$ $2^{-x}=\frac{100}{g(x)}-1\implies$ $-x=\log_2(\frac{100}{g(x)}-1)\implies$ $x=-\log_2(\frac{100}{g(x)}-1)$


0

Replace sequences with functions: $$ N_{t} \longrightarrow N(t)$$ Consider $R$ as rate of population growth: $$R(t)=N'(t)$$ If $$N(t+1)=N(t) exp(r(1-\frac{N(t)}{k}))$$ Then for small $h \approx 0$ you can linearize the equation: $$N(t+h)=N(t)+hN(t) exp(r(1-\frac{N(t)}{k}))$$ Then simply: $$R(t)=N'(t)=\frac{N(t+h)-N(t)}{h}=N(t) ...


0

If we want to define a function $f \colon \mathbb{R} \to \mathbb{C}$, $f(t) = e^{it}$ in a way that meshes with our formulas for the real exponential function, it makes sense to require that $f(0) = 1$ and $f'(t) = if(t)$. Let's prove that we must then have $f(t) = \cos t + i\sin t$. Write $f(t) = x(t) + iy(t)$. We have $f'(t) = x'(t) + iy'(t)$. First note ...


2

The solutions are counting multiplicity and include complex solutions. It's called the fundamental theorem of algebra. Low-level proofs are not easy to come by, however.


0

It was already mentioned in comments and other answers that the inequality $\exp(x)>1$ for $x>0$ is easy. Here is rather elementary proof that $\exp(x)\le1$ for $x<0$. Let us denote $$f_n(x)=1+x+\frac{x^2}{2!}+\dots+\frac{x^n}{n!}.$$ Notice that $f_{n+1}'(x)=f_n(x)$. Try to prove by induction that for $n=1,2,\dots$ that: $f_{2n-1}(x)$ is ...


0

the first part $e^x > 1$ for $x > 0$ follows from $e^x = 1 + x + \frac{x^2}{2} + \cdots$ because $x > 0$ implies $x^2 > 0, x^3 > 0, \cdots$ the second part $0 < e^x < 1$ for $x < 0$ does not seem to be elementary. if you can establish even an instance of the additive property $e^{-x} = \dfrac{1}{e^x},$ then you are done. one way of ...


1

If $x>0$, you have a series with positive terms, so its sum is greater than the sum of the first two terms, which is $1+x$ and $1+x>1$. If $x<0$ you have an alternating series. It's a theorem that for alternating series, the error bound when you take the sum up to rank $m$: $\sum_{k=0}^m a_k$, is at most $\lvert a_{m+1}\rvert$ and the error has the ...


3

The power series of the exponential function is defined on $\Bbb R$ so we can differentiate it term by term on $\Bbb R$ and we get $$\exp'(x)=\exp(x)$$ Moreover, we see easily that $\exp(x)>0$ for $x\ge0$ and using the Cauchy product we get $$\exp(x)\exp(y)=\exp(x+y),\quad \forall (x,y)\in\Bbb R^2$$ hence $$\exp(-x)\exp(x)=\exp(0)=1\implies ...


1

For $x>0$, $$\exp(x)=1+x+\frac{x}{2}+\cdots>1+0+0+\cdots=1$$ Since $\exp(0)=1$ and $\exp$ is strictly monontone increasing, $\exp(x)<1$ for $x<0$.


1

Try using logarithms. A logarithm is defined as follows, If $a^{b}=c$, then $\log_{a} c=b$ So, similarly, here we get $m=\log_{d} n$. We can further simplify it by changing the logarithms' base to $10$, $m=\frac{\log_{10} n}{\log_{10} d}$ If you want to learn more about logarithms, go here


3

Using any logarithm $\log$, we have $$\log n = \log (d^m) = m \log d,$$ so $$m = \frac{\log n}{\log d} = \log_d n.$$


0

\begin{align} 0&=\frac{4}{3}e^{3x}+2e^{2x}-8e^x\tag{1} \\[1em] & = \frac{\frac{4}{3}e^{3x}+2e^{2x}-8e^{x}}{2e^x}\tag{2} \\[1em] & = \frac{2}{3}e^{2x}+e^x-4\tag{3} \\[1em] \end{align} Now let $\xi=e^x,\therefore e^{2x}=\left(e^x\right)^2=\xi^2.$ This gives us \begin{align} 0&=\frac{2}{3}\xi^2 +\xi-4\tag{4} \\[1em] \therefore \xi & = ...


3

Hint: The $x$-intercept is when $\frac43 e^{3x}+2e^{2x}-8e^{x} = 0$. Now set $y = e^{x}$, so your equation is $$\frac{4}{3}y^3+2y^2-8y = 0 $$ which means $$y\left(\frac{4}{3}y^2+2y-8\right) = 0. $$


7

This equation cannot be solved using “traditional” algebraic manipulations. In this case, one would use the Lambert W function: $$W(x): x = W(x)\cdot e^{W(x)}$$ or in other words, it is the solution of the equation $x = w e^w$. With this knowledge, we can try to substitute $y:=\frac{1}{x}$: $$\Rightarrow 0 = e^y-\frac{1}{y} \Rightarrow \frac{1}{y} = e^y ...


0

(i) We need to solve $$\log y = x^2 - x + 6 \, \Leftrightarrow \, x^2 - x + 6 - \log y = 0.$$ Use the quadratic formula with $a = 1, \, b = -1, \, c = 6 - \log y$. (ii) For any function $f$, and its inverse $f^{-1}$, we have $\operatorname{dom} f^{-1} = \operatorname{range} f$. This makes sense because the definition of an inverse function is $f^{-1}(f(x)) ...


0

In reference to Axoren, the following is true for (1): \begin{align} 9^{x+1}=27^{2x-3} & \implies\left(3^2\right)^{x+1}=\left(3^3\right)^{2x-3}\tag{1}\\[1em] & \implies \left(3\right)^{2x+2}=\left(3\right)^{6x-9}\tag{2} \\[1em] & \implies \left(2x+2\right)\log_3\left|3\right|=\left(6x-9\right)\log_3\left|3\right|\tag{3}\\[1em] & \implies ...


1

Hints: $$\log(a^b) = b \log(a) \\ \log(ab) = \log(a) + \log(b) $$ Example for part $c)$: $$ 210 = 40(1.5)^x \\ \log(210) = \log(40 (1.5)^x) \\ \log(210) = x\log(1.5) + \log(40) \\ \frac{\log(210) - \log(40)}{\log(1.5)} = x \\ x \approx 4.0897 $$


1

I had decided to solve this problem using $y=i(1-a)^t$ to solve this problem, where $y$ is the final amount, $i$ is the initial amount, $a$ is the rate of decrease or "cooling factor" in this case, and $t$ is the number of time periods. you will see why I used this formula in a little bit. a) To solve for $a$, we can input some values from the table to fill ...


5

I think this is basically a different way to state exactly what André Nicolas said in a comment, but observe that the largest power of a given prime $p$ found as a factor in a number less than $n$ is explicitly given by $$ \lfloor\log_p(n)\rfloor=\left\lfloor\frac{\ln n}{\ln p}\right\rfloor $$ so we can write $LCM(n)=LCM(1,2,...,n)$ explicitly as $$ ...


1

Let the initial volume of the container be $V_0$ and the density be $\rho$. Let the evaporation and condensation be uniform and that 2.1% of the volume is lost everytime the purifying process is over. Thus the model is $${\rho\times(\dot V_0 - \dot V_1)} = 0.021*\rho\times\dot V_0$$ Cancelling $\rho$, and converting the volumetric rate to volume, You ...


1

After our comment conversation, we see that the equation would be $$\text{amount}=\text{initial}(0.979)^x$$ And to see how many cycles it takes to get to half the initial amount would be $$\frac{c}{2}=c(0.979)^x\\\frac{1}{2}=0.979^x$$


2

The Poisson models the number of arrivals in a certain fixed time. It is a discrete distribution, taking on values $0,1,2,\dots$. The exponential models the waiting time between consecutive arrivals. It is a continuous distribution. There is a connection, since they are used in modelling two different features of the same phenomenon. But they are quite ...



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