Rewrite Each Equation In Exponential: Fill & Download for Free

What common base can be used to rewrite each side of the equation 2^x+3-3=5?Log base 2.You are trying to undo an exponential power. Logarithm is the inverse of exponential power. Since your exponential power is base 2, you would use log base 2 to undo it.I am assuming the problem looks like:And Not like this:Because these two options the -3+3 just cancel each other out immediately, so it seems like a waste of ink to even write.

We can rewrite this homogeneous linear ordinary differential equation as[math]\displaystyle (\frac{d^2}{dx^2} - 3 \frac{d}{dx} [/math][math][/math][math]+ 2)u = 0[/math]So either it is the trivial solution [math]u \equiv 0[/math] or [math]u[/math] is is some linear combination of eigenfunctions of the derivative operator.The exponential function is precisely the sort of thing we are looking for[math]\displaystyle \frac{d}{dx} e^{\lambda x} = \lambda e^{\lambda x}[/math]So now if [math]u[/math] is an exponential function the above simplifies to[math]\lambda^2 - 3 \lambda [/math][math][/math][math]+ 2 = 0[/math]We can factor and we find that [math]\lambda = 1, 2[/math]So the solutions to the ODE are of the form[math]u = C_1 e^x [/math][math][/math][math]+ C_2 e^{2x}[/math]We can check this solution by plugging in each of the basis solutions back in.[math]e^x - 3 e^x [/math][math][/math][math]+ 2 e^x = 0[/math][math]4 e^{2x} - 3(2)e^{2x} [/math][math][/math][math]+ 2e^{2x} = 0[/math]

Expanding on Anon's answer a little.You can describe a wave by playing around with trig functions a little. For example, you can describe your wavefunction as a cosine wave with some amplitude [math]A[/math] and phase factor [math]\phi[/math]:[math]\psi = A \cos(\omega t - \phi)[/math]where [math]\omega[/math] is just the [angular] frequency in time. The phase factor can tell you a few things. One thing you might use it for is to offset the entire wave at [math]t=0[/math] let's say, in case [math]\psi[/math] doesn't start at [math]A[/math]. Another thing you can do is make [math]\phi[/math] a function of space,[math]\psi = A \cos(\omega t - kx)[/math]where the wavenumber [math]k = 2\pi/\lambda[/math], which means that [math]\psi[/math] is moving in the positive [math]x[/math]-direction. You can replace the wavenumber and spatial variable with vectors, getting [math]\mathbf{k}\cdot \mathbf{r}[/math].There's a simple way to describe waves, and it's by thinking about a Phasor, which is just a vector that describes a wavefunction. This animation from the Wikipedia page explains it quite well ([math]\psi[/math] is [math]y[/math] here):So [math]A[/math] is just the radius of this circle, and the phase factor just tells you where to start. The vertical axis of this circle is actually the complex plane (the horizontal being the real one of course). In this way, we can reformulate wavefunctions from using just trig functions to a description that includes complex numbers. If you want something more mathematical, continue on (the rest is taken from some notes I had of something I found online, if I find the page I'll link it here). WRT what Anon was talking about, the link to that is at the end.First, you can derive something that looks like[math]y = A \cos\omega t \cdot \cos\phi [/math][math][/math][math]+ A \sin\omega t \cdot \sin\phi [/math][math][/math][math][/math]and noticing that [math]A \cos\phi[/math] (or the sine version) describe a vector in the complex plane ([math]A[/math] is the magnitude, and [math]\cos\phi[/math] tells you the angle made with the horizontal), you can group those together to make your "phasor", so you get [math]\mathbf{A} = A\cos\phi [/math][math][/math][math]+ iA\sin\phi = Ae^{i\phi}[/math] using Euler's equation.Then if you rewrite the sum of cosines and sines equation in terms of exponentials also using Euler's equation, you'll get[math]\cos\omega t = \frac{e^{i\omega t} [/math][math][/math][math]+ e^{-i\omega t}}{2}[/math]and[math]\sin\omega t = \frac{e^{i\omega t} - e^{-i\omega t}}{2i}[/math]so finally writing the whole wavefunction gets us[math]\psi = \frac{1}{2}(\mathbf{A}^* e^{i\omega t} [/math][math][/math][math]+ \mathbf{A}e^{-i \omega t})[/math]where [math]\mathbf{A}^*[/math] is just the complex conjugate of [math]\mathbf{A}[/math] (i.e. flip signs in front of all imaginary numbers).By the way, the [math]\mathbf{A}[/math]'s are the complex wave amplitudes that Feynman talks about in his infamous lectures if you have read them (which you should, now that you have a decent idea of what these amplitudes actually are). If you think about how waves interfere with one another, it all happens with these phasor things. They are at least very strongly analogous to your state vector (I think saying they are the same might be an abuse of terms). You'll notice that the complex parts of these waves will cancel out when interfering with each other, i.e. what you'll get is a real number equal to the real part of [math]\mathbf{A}[/math]. So you can write your WF like,[math]\psi = \mathbf{A}e^{-i(\omega t - \mathbf{k} \cdot \mathbf{r})}[/math]I think now Anon's post is clear since you can see how it's nicer to apply arbitrary phase in this formulation--all you do is multiply the whole thing with your new phase, i.e. [math]\psi e^{i\theta}[/math]. The phase is not measurable except in an indirect way, like when two waves with different wave amplitudes interact, and then you get a value that is not equal to either amplitudes. There are some guarantees on how you can construct and manipulate wavefunctions so that when you make a measurement of these complex wavefunctions (which can also be thought of as a projection of your state vector onto another basis vector that corresponds to the way you're measuring), you get a real number.