Solving Equations Involving Exponentials and Logarithms

[ProPM]

Find the values of a and k if the graph with equation f(x) = ae^-kx passes through the points (1, e) and (-1,2e)

So, from the information above I managed to derive two equations to solve simultaneously:

ae^-k = e
ae^k = 2e

I am pretty sure those are correct, but I am not 100%. I just think I am having more of a problem solving them!

Here is what I tried:

I tried dividing the second equation by the first:

therefore: e^k / e^-k = e^2k
2e / e = 2, hence:

e^2k = 2

To solve for k I did:

(e^k)² = 2 and I let e^k = y, hence
y = √2

e^k = √2
k log e = 1/2 log 2
k = 1/2 log 2 / log e

I hope my maths is not very confusing, although I do recognized there must be other shorter paths...

Thanks

 

[rock.freak667]

that is correct but from here

e^2k = 2

you could have taken advantage of the 'ln' function to get a simpler looking answer of

2k =ln2 such that k = ½ ln2.

 

[ProPM]

Cool, thanks rock freak

 

[ProPM]

Oh, if you wouldn't mind, I have a series of question in the following format and I wanted to make sure I am heading down the right path: Consider f: x: e^x+1 - 3

a) Find the inverse function
b) State the range and domain

a) I did: e^x+1 = y + 3

log_e y + 3 = x + 1
log_e (x+3) - 1 = y

For the range, plotting the graphs in my calculator I figured:
For f: Domain: x goes all the way to infinity
Range: y is always bigger than - 3

For f^-1: Domain x bigger than -3 and range would be all the way to infinity.

I think I am good with that but I am a bit intimidated when ln is involved, since I am not used working with it, for example: ln (x-1) + 2

Thanks once again.

 

[Mark44]

Oh, if you wouldn't mind, I have a series of question in the following format and I wanted to make sure I am heading down the right path: Consider f: x: e^x+1 - 3

a) Find the inverse function
b) State the range and domain

a) I did: e^x+1 = y + 3

log_e y + 3 = x + 1
log_e (x+3) - 1 = y

Looks fine, but you can write ln instead of log_e.

For the range, plotting the graphs in my calculator I figured:
For f: Domain: x goes all the way to infinity

From where? The domain of f is all real numbers, or (-inf, +inf).

Range: y is always bigger than - 3

For f^-1: Domain x bigger than -3 and range would be all the way to infinity.

See above. "All the way to infinity" is not precise. Just say all real numbers.

I think I am good with that but I am a bit intimidated when ln is involved, since I am not used working with it, for example: ln (x-1) + 2

Thanks once again.

 

[ProPM]

Ok, thanks!

As for my ln dilema, I will be more precise: ln x - 4 where x is bigger than 0.

so: ln x = y + 4

hence, e^y+4 = x

Is that ok for the inverse?

Thanks

 

[Mark44]

Ok, thanks!

As for my ln dilema, I will be more precise: ln x - 4 where x is bigger than 0.

so: ln x = y + 4

hence, e^y+4 = x

Is that ok for the inverse?

Thanks

No.
I'm assuming that you're still working with y = e^x+1 - 3, and solving the equation for x.

y = e^x+1 - 3
<==> y + 3 = e^x+1
<==> ln(y + 3) = x + 1
<==> ln(y + 3) - 1 = x

The two equations y = e^x+1 - 3 and x = ln(y + 3) - 1 are equivalent, which means that any pair of numbers (x, y) that is a solution to one equation is also a solution to the other equation. This also means that the graphs of the two equations are exactly the same. The only difference is that one equation gives y as a function of x, and the other equation gives x as a (different) function of y.

If y = f(x) = e^x+1 - 3, then
x = f^-1(y) = ln(y + 3) - 1

To write f^-1 as a function of x, replace x for y and y for x in the 2nd equation above.

IOW, y = f^-1(x) = ln(x + 3) - 1