Solving Equations Involving Exponentials and Logarithms

In summary, the inverse function of f(x) = ex+1 - 3 is f-1(x) = ln(x + 3) - 1. The domain of f is all real numbers, and the range is all real numbers greater than -3. The domain of f-1 is all real numbers greater than -3, and the range is all real numbers.
  • #1
ProPM
66
0
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
aek = 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: ek / e-k = e2k
2e / e = 2, hence:

e2k = 2

To solve for k I did:

(ek)2 = 2 and I let ek = y, hence
y = √2

ek = √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
 
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  • #2
that is correct but from here

e2k = 2

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

2k =ln2 such that k = ½ ln2.
 
  • #3
Cool, thanks rock freak :smile:
 
  • #4
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: ex+1 - 3

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

a) I did: ex+1 = y + 3

loge y + 3 = x + 1
loge (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.
 
  • #5
ProPM said:
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: ex+1 - 3

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

a) I did: ex+1 = y + 3

loge y + 3 = x + 1
loge (x+3) - 1 = y
Looks fine, but you can write ln instead of loge.
ProPM said:
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).
ProPM said:
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.
ProPM said:
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.
 
  • #6
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, ey+4 = x

Is that ok for the inverse?

Thanks
 
Last edited:
  • #7
ProPM said:
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, ey+4 = x

Is that ok for the inverse?

Thanks

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

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

The two equations y = ex+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) = ex+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
 

1. How do I plot a graph of a function?

To plot a graph of a function, you will need to first identify the independent and dependent variables. Then, choose a range of values for the independent variable and use the function to calculate the corresponding values for the dependent variable. Plot these points on a coordinate plane and connect them with a smooth curve.

2. What is the difference between linear and non-linear functions?

A linear function is a function where the output (dependent variable) changes at a constant rate as the input (independent variable) changes. This results in a straight line when plotted on a graph. Non-linear functions, on the other hand, have a variable rate of change and produce curved lines on a graph.

3. How do I determine the domain and range of a function?

The domain of a function is the set of all possible values for the independent variable. To determine the domain, look for any restrictions on the independent variable, such as square roots or division by zero. The range of a function is the set of all possible values for the dependent variable. To determine the range, look at the output values of the function.

4. What is the difference between a function and an equation?

A function is a relation between two sets of variables, where each input (x-value) has only one output (y-value). An equation, on the other hand, is a mathematical statement that shows the relationship between two or more variables. An equation can represent a function, but not all equations are functions.

5. How do I determine the x and y-intercepts of a function?

The x-intercept of a function is the point where the graph crosses the x-axis, and the y-intercept is the point where the graph crosses the y-axis. To find the x-intercept, set the y-value to 0 and solve for x. To find the y-intercept, set the x-value to 0 and solve for y. These points can then be plotted on the graph.

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