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How do you find the exponential model y=ae^(bx) that …socratic.org › … › Population ModelsSince x=0 in the first equation I know that the exponent will be 0 and anything raised to 0 is evaluates to 1 so let's do this first to easily solve for a. 4=ae^(b0) 4=a×1 Having the value of a we use the second equation to solve for b. 1/2=4e^(b5) 1/8=e^(b5) ln(1/8)=b5 ln(1/8)/5=b ... Socratic Meta Featured Answers Precalculus . Science ...

Since x=0 in the first equation I know that the exponent will be 0 and anything raised to 0 is evaluates to 1 so let's do this first to easily solve for a. 4=ae^(b0) 4=a×1 Having the value of a we use the second equation to solve for b. 1/2=4e^(b5) 1/8=e^(b5) ln(1/8)=b5 ln(1/8)/5=b ... Socratic Meta Featured Answers Precalculus . Science ...
How do you find the exponential model y=ae^(bx) th...

How do i solve $e^{ax}-e^{bx}=c$ for $x$? - Stack …https://math.stackexchange.com/.../how-do-i-solve-eax-ebx-c-for-xHow do i solve $e^{ax}-e^{bx}=c$ for $x$? The constants $a$, $b$ and $c$ are real numbers. It is the final form of a longer equation that I simplyfied. Edit: The ...As others have said, you are in for a numeric solution, but it can be made simpler. It looks like you measure $t$ in hours, so it is in the range $10-30$, which means $ax$ is in the range $-.1$ to $-.3$ and $bx$ is in the range $-1$ to $-3$. I would rewrite the equation as $x=\frac 1a \log (c+e^{bx})$ and iterate, starting with $x_0=20$. It should converge quickly.Best answer · 4I'll simply reiterate what others have said: There is no general way to solve equations of the form you have. There are a handful of values of $a, b$ for which a solution to $\;e^{ax} - e^{bx} = c\;$ can always be solved: e.g., if $a, b \in \{1, 2, 3, 4\}$, you're in luck. (Of course, if $c = 0$, you're also in luck.) There are numerical methods for solving such an equation for given values of $a, b, c$. If you have only the general form you posted, I'm afraid there's no all-encompassing solution. But if you have particular values of $a, b, c$ in mind, feel free to edit the post and let us have a look at those values.12Declaring $y=e^x$ gives $$y^a-y^b=c$$ For large $a$ and $b$ this equation is not solvable: it's either a polynomial of high degree or some other complicated beast. Numerics are the way to go here.9Write it as $(e^x)^a$ and the same for the second one too. Now check if $c$ is positive or negative. If +ve, take $(e^x)^b$ common. Now the left thing $((e^x)^a)-1$ is not divisible by $e^b$ anymore. So divide $c$ by $e^b$ (if $c$ has a factor). So the no. of times you can divide it by it is $x$. Just put $x$ afterwards and surely check. Just visualize it and then do it. You can use it to solve equations like $(2^m)-(2^n)=56$. Actually someone gave me the equation above (the one with powers of $2$) and then asked me to find $m$ and $n$. I divided $56$ by $2$ as many times as i can until i reached $7$ ($3$ times). So $n$ is $3$ and $m$ is $6$. Surely try this too!2After you edited, your equation is: $$ e^{-\lambda_M t} - e^{-\lambda_T t} = \frac{A_T(\lambda_T−\lambda_M)}{A_M\lambda_T}. $$ Now you want to get a reasonably accurate $t$ by numerics to approximate the true time. Here is my suggestion: due to the derivative of the left hand side decays very fast to zero, common root finding algorithms like Newton's method (or any other root finding method belonging in the class of Householder's methods) may not converge very well. You may wanna try the old school Bisection method .0

How do i solve $e^{ax}-e^{bx}=c$ for $x$? The constants $a$, $b$ and $c$ are real numbers. It is the final form of a longer equation that I simplyfied. Edit: The ...
math.stackexchange.com/.../how-do-i-solve-eax-ebx-...

a a + bx

a exp(-bx) + c f = a e-bx + c 0 0.2 0.4 0.6 0.8 1 0 2 4 6 8 10x a exp(-bx) + c Example: Chemical kinetics first order decay of a reactant towards equilibrium.
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Exponential and Logarithmic Modelsdl.uncw.edu/digilib/Mathematics/Algebra/mat111hb/EandL/elmodels/el...Exponential Growth: y = a e bx, b > 0. Exponential Decay: y = a e-bx, b > 0. Example 1. During the 1980s the population of a certain city went from 100,000 to 205,000. Populations by year are listed in the table below. Year 1980

Exponential Growth: y = a e bx, b > 0. Exponential Decay: y = a e-bx, b > 0. Example 1. During the 1980s the population of a certain city went from 100,000 to 205,000. Populations by year are listed in the table below. Year 1980
dl.uncw.edu/digilib/Mathematics/Algebra/mat111hb/E...

Exponential Functions Ae^(bx) - YouTubehttps://www.youtube.com/watch?v=mcRB-_eWha8Apr 17, 2013 · YouTube Premium Loading... Get YouTube without the ads. Working... No thanks 1 month free. Find out why Close. Exponential …

Apr 17, 2013 · YouTube Premium Loading... Get YouTube without the ads. Working... No thanks 1 month free. Find out why Close. Exponential …
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SOLUTION: Solve the equation for x. e^(ax) = Ce^(bx ...https://www.algebra.com/algebra/homework/logarithm/logarithm.faq...Algebra -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve the equation for x. e^(ax) = Ce^(bx), where a is not equal to …

Algebra -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve the equation for x. e^(ax) = Ce^(bx), where a is not equal to …
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How do you find the exponential model y = ae^(bx) that ...socratic.org › … › Exponential and Logistic GraphsIt is nice that we are given the point, (0,8), because it allows us to find the value of a before we find the value of b: Substitute the point (0,8) into y=ae^(bx): 8=ae^(b(0)) Any number raised to the zero power is 1: 8 = a(1) a = 8 Use the point, (1,3), to find the value of b: 3 = 8e^(b(1)) e^b= 3/8 b = ln(3/8) The final equation is: y = 8e^(ln(3/8)x) Often, the …

It is nice that we are given the point, (0,8), because it allows us to find the value of a before we find the value of b: Substitute the point (0,8) into y=ae^(bx): 8=ae^(b(0)) Any number raised to the zero power is 1: 8 = a(1) a = 8 Use the point, (1,3), to find the value of b: 3 = 8e^(b(1)) e^b= 3/8 b = ln(3/8) The final equation is: y = 8e^(ln(3/8)x) Often, the …
How do you find the exponential model y = ae^(bx) ...

jwilson.coe.uga.edujwilson.coe.uga.edu/EMAT6680Fa08/KimH/assignment1hjk/test3.htmlIn addition, y = - a e bx is symmetric with y = a e bx with respect to the x-axis, so is a horizontal reflection of y = a e bx. Lastly, let’s observe the effects of parameter c on the graph of the exponential function y = a e bx + c.

In addition, y = - a e bx is symmetric with y = a e bx with respect to the x-axis, so is a horizontal reflection of y = a e bx. Lastly, let’s observe the effects of parameter c on the graph of the exponential function y = a e bx + c.
jwilson.coe.uga.edu/EMAT6680Fa08/KimH/assignment1h...

Review - University Of Illinoishttps://mste.illinois.edu/malcz/ExpFit/REVIEW.htmlExponential Review. Base e. The number e has been called one of the most important numbers in all of mathematics. However, it is important to remember that e is just a number. Calculated to nine decimal places, e = 2.718281828. ... y = a e^(b x) where a and b are constants. The curve that we use to fit data sets is in this form so it is ...

Exponential Review. Base e. The number e has been called one of the most important numbers in all of mathematics. However, it is important to remember that e is just a number. Calculated to nine decimal places, e = 2.718281828. ... y = a e^(b x) where a and b are constants. The curve that we use to fit data sets is in this form so it is ...
mste.illinois.edu/malcz/ExpFit/REVIEW.html