Question

How do you solve -5e^(-4x+2)+3=1/2log(x^2+1)? Please help, thank you.

Answer 1

\[\LARGE -5e^{-4x+2}+3=\frac{ 1 }{ 2 }\log(x^2+1)\]
Is that you equation?

Answer 2

There are various methods, but you will not find one that leads to a articular pretty result. First, are you sure there IS a solution?

Answer 3

Let me know if you still need help, @bifurious37

Answer 4

Looks to me like there are two solutions.
One around x = 3/4 and one around x =20.
Happy hunting!

Answer 5

I'm pretty sure there is a solution(s) because the question asks me to use a calculator to solve this equation, and then to round my answer to three decimals. Could you explain to me how you found those solutions @tkhunny

Answer 6

Yes, that is my equation. Could you help me solve it? @3mar

Answer 7

Your equation is\[\LARGE -5e^{-4x+2}+3=\frac{ 1 }{ 2 }\log(x^2+1)\]

Answer 8

One way in which you could solve this equation with a calculator would be as follows:
1) graph the left side of the equation
2) on the same set of coordinate axes, graph the right side of the equation
3) approximate the coordinates of any point at which the 2 separate graphs meet.

If you're in calculus, you could use Newton's Method for Approximating Roots to find increasingly accurate solutions.

Please let me know whether or not this is clear and whether or not you need further help.

Answer 9

Thank you so much! @mathmale

Answer 10

my great pleasure!

Answer 11

Another interesting way is to just solve for ONE of the x's. I did this:

\(f(x) = \dfrac{\ln\left(\dfrac{\dfrac{1}{2}\ln\left(x^{2}+1\right) - 3}{-5}\right)-2}{-4}\)

Then, since I already know there was a solution around x = 0.75, I started with that, calculating f(0.75) = 0.647029599598176. Usinng that value, I calculated f(0.647029599598176) = 0.642719281723003. and etc.

These are my results with 10 iterations:

0.75
0.647029599598176
0.642719281723003
0.642545617527228
0.642538633128101
0.642538352251128
0.64253834095572
0.642538340501477
0.64253834048321
0.642538340482475
0.642538340482446

You can see that this seems to be converging. It's not super quick,but it is stable for this one. It can be a good way to go.

Interestingly, with flatter derivatives, we are not so lucky around x = 20

20
2.35304317424737
0.721528273756258
0.645809073367672
0.642670045505043
0.642543637253352
0.642538553491407
0.642538349048555
0.642538340826929
0.642538340496298
0.642538340483002

It seems to want to find the one we already know. Something more sophisticated may be required.

Fun little exploration.

Answer 12

Resorted to Newton's Method for the other one. The deal here is that if it converges, you get QUADRATIC convergence.

So, I created: \(f(x) = \dfrac{1}{2}\ln\left(x^{2}+1\right) - \left(3 - 5e^{2-4x}\right)\)

Then the usual, \(x_{0} = 20\) and \(x_{i+1} = x_{i} - \dfrac{f(x_{i})}{f'(x_{i})}\)

This leads to:

20
20.0605366912516
20.0606279434231
20.0606279436297
20.0606279436297

Like I said, that was quicker.

Starting with \(x_{0} = 0.75\) gives

0.75
0.613693618956242
0.640867410057331
0.642532499583752
0.642538340410892
0.642538340482445

Very fast. However, if it doesn't converge, speed isn't very helpful. In this case, both converged.

It should strike you as at least a little amazing that these two methods, with entirely different-looking functions, managed to find the same solution.