What is e ln x

e function

The exponential function belongs to the group of exponential functions and is also called the "natural exponential function". In order to understand the e-function, we will look at all the topics in this article that you need for the calculation with the e-function.

Table of Contents


Basics of the exponential function

A function is called an exponential function (to base b) if it has the form

\ begin {align *}
f (x) = b ^ x,
\ end {align *}

where b denotes any positive constant.

If b = e, one generally speaks of “the” e-function.

Please don't get confused by this "e". This is Euler's number - a perfectly normal number e = 2.718281828459045235 ... The form of the exponential function reminds us of that of the power expression, with the role of base and exponent being reversed!

So here we cannot derive as usual and have to rewrite the expression for derivation purposes.

The following applies:
\ begin {align *}
b ^ x = e ^ {\ ln (b) \ cdot x}
\ end {align *}

For the case that b = e, the following applies as a consequence of the power laws for the e-function:

\ begin {align *}
e ^ 0 = 1, \ \ e ^ 1 = e, \ \ e ^ x \ cdot e ^ y = e ^ {x + y}
\ end {align *}

Here you can see the graph of the exponential function

As you can see, the graph of the exponential function always runs above the x-axis. The graph approaches the x-axis, but will not intersect it. This in turn means that the classical exponential function has no zeros.

The strictly monotonically increasing course of the function intersects the y-axis at the point (0 | 1).

Not yet understood the subject of e-function? Check out Daniel's introduction to the subject!

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Calculating with the e-function

To solve e-functions, one usually uses their inverse function, the natural logarithm ln.

A useful context is

\ begin {align *}
e ^ {\ ln (x)} = x \ quad \ textrm {or} \ quad \ ln (e ^ x) = x.
\ end {align *}

Pay attention to the logarithmic laws! Some follow Examples to solve e-functions:

\ begin {align *}
e ^ {2x} \ cdot (x ^ 2-2) = 0 \
e ^ {2x} = 0 \ \ vee \ x ^ 2-2 & = 0 \ quad | +2 \
x ^ 2 & = 2 \ quad | \ sqrt {~~} \
x_1 = \ sqrt {2} & \ wedge x_2 = - \ sqrt {2}
\ end {align *}

Why does $ e ^ {2x} = 0 $ not provide a solution? If one logarithmizes both sides it follows that $ \ ln (2x) = \ ln (0) $. Since the natural logarithm is not defined for 0 ($ D = (0, \ infty)) $, there is no solution.

Examples

\ begin {align *}
1. \ quad 8e ^ {- 2x} -16 & = 0 \ quad \ quad \ quad \ \ mid + 16 \
8e ^ {- 2x} & = 16 \ quad \ quad \ \ \ mid: 8 \
e ^ {- 2x} & = 2 \ quad \ quad \ \ quad | \ ln \
\ ln (e ^ {- 2x}) & = \ ln (2) \
-2 x & = \ ln (2) \ quad \ quad |: (- 2) \
x & = - \ ln (2) / 2
\ end {align *}

\ begin {align *}
2. \ quad 4e ^ {3x} -e ^ {2x} & = 0 \ quad \ quad \ quad | + e ^ {2x} \
4e ^ {3x} & = e ^ {2x} \ quad \ quad \ | \ ln \
\ ln (4 \ cdot e ^ {3x}) & = \ ln (e ^ {2x}) \
\ ln (4) + \ ln (e ^ {3x}) & = 2x \
\ ln (4) + 3x & = 2x \
\ ln (4) & = - x \
- \ ln (4) & = x
\ end {align *}

Check out the full exponential function playlist to review!

Solve equations at e ^ x, overview 1, e-function | Math by Daniel Jung

Deriving the exponential function

An e-function is derived as follows: You use the chain rule “officially”, but it is actually a lot easier. We consider the function for this

\ begin {align *}
f (x) = e ^ {5x},
\ end {align *}

which we want to derive with respect to x. To do this, we simply write the term with the exponential function again and multiply the thing with the derived exponent. The exponent here is 5x and derived that would be simply 5. Then follows for the derivative

\ begin {align *}
f '(x) = e ^ {5x} \ cdot 5.
\ end {align *}

Another example

$
\ begin {array} {c | c}
f (x) & f '(x) \ \ hline
e ^ x & e ^ x \ \ hline
2e ^ x & 2e ^ x \
3e ^ x & 3e ^ x \ \ hline
e ^ {2x} & 2e ^ {2x} \
e ^ {3x} & 3e ^ {3x} \
e ^ {x ^ 2} & 2xe ^ {x ^ 2} \
e ^ {2-4x} & -4e ^ {2-4x} \ \ hline
20e ^ {3x} & 3 \ cdot 20 e ^ {3x} \
x \ cdot e ^ {2x} & product rule \
\ end {array}
$

If an exponential function is multiplied by other functions, we have to apply the already known product rule.

You can see how this works again in this video!

Derive e-function in the product, product and chain rule, derivation of exponential function

Math Abi'21 study booklet incl. Exercise collection

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Another example

\ begin {align *}
f (x) & = \ underbrace {(x ^ 2-2)} _ {\ text {u (x)}} \ cdot \ underbrace {e ^ {- 2x}} _ {\ text {v (x)} } \
\ textrm {with} \ quad u (x) & = x ^ 2-2 \ quad u '(x) = 2x \
\ textrm {and} \ quad v (x) & = e ^ {- 2x} \ quad \ quad v '(x) = -2e ^ {- 2x}
\ end {align *}

Thus for the first derivation we get:

\ begin {align *}
f '(x) = 2xe ^ {- 2x} + (x ^ 2-2) \ cdot (-2e ^ {- 2x})
\ end {align *}

If you want, you can now rewrite this expression:

\ begin {align *}
f '(x) & = e ^ {- 2x} (2x + (x ^ 2-2) (- 2)) \
& = e ^ {- 2x} (2x-2x ^ 2 + 4) \
& = e ^ {- 2x} (- 2x ^ 2 + 2x + 4)
\ end {align *}

Daniel shows you in the video how you can derive the exponential function

Derive exponential function, derivation exponential function, simple overview | Math by Daniel Jung

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Integrate the e-function

\ begin {array} {c | c}
f (x) & F (x) \ \ hline
e ^ x & e ^ x \
e ^ {2x} & \ frac {1} {2} e ^ {2x} \
e ^ {3x} & \ frac {1} {3} e ^ {3x} \
e ^ {4-2x} & \ frac {-1} {2} e ^ {4-2x} \
20e ^ {10x} & \ frac {20} {10} e ^ {10x} \
3e ^ {5-2x} & \ frac {3} {- 2} e ^ {5-2x} \
e ^ {x ^ 2}, e ^ {x ^ 3} & \ textrm {Not possible!} \ \
2x \ cdot e ^ {- 2x} & \ textrm {partial integration} \
2x \ cdot e ^ {x ^ 2} & \ textrm {Substitution} \
\ end {array}

Regardless of whether you want to determine zeros, form derivatives or antiderivatives: Pay attention to the structure of the function! Is there just a sum or difference, is a product of a term with a variable times e to the power of something to be recognized?

For more in-depth information, watch Daniel's video on the topic of antiderivatives in e-function.

Antiderivative e ^ x overview, e-function, integration options | Math by Daniel Jung

Symmetry of the exponential function

Is $ f (x) = x ^ 2 \ cdot e ^ {- x ^ 2} $ axially symmetric to the y-axis? Then the following should apply:

\ begin {align *}
f (-x) & = f (x) \
(-x) ^ 2 \ cdot e ^ {- (- x) ^ 2} & = x ^ 2 \ cdot e ^ {- x ^ 2} \
x ^ 2 \ cdot e ^ {- x ^ 2} & = x ^ 2 \ cdot e ^ {- x ^ 2} \
\ end {align *}

Is $ f (x) = - 10x \ cdot e ^ {x ^ 2} $ point symmetric to the origin? Then the following should apply:

\ begin {align *}
f (-x) & = - f (x) \
-10 \ cdot (-x) \ cdot e ^ {(- x) ^ 2} & = - \ left (-10x \ cdot e ^ {x ^ 2} \ right) \
10 x \ cdot e ^ {x ^ 2} & = 10x \ cdot e ^ {x ^ 2} \
\ end {align *}

Check out Daniel's tutorial on symmetry.

Symmetry in e-functions, exponential function, math help online | Math by Daniel Jung

Limit behavior of the e-function

Exponential functions and their graphs are examined in the same way as completely rational functions. Just the behavior
an exponential function for $ x \ to + \ infty $ and for $ x \ to - \ infty $ is governed by other rules.

  • For $ x \ to + \ infty $, $ e strives for ^ x \ to + \ infty $.
  • For $ x \ to - \ infty $, $ e ^ x \ to 0 $ strives, i.e. the x-axis is the asymptote of the graph of f with $ f (x) = e ^ x $.

In addition, for $ n \ geq 1 $:

  • For $ x \ to + \ infty $, $ x ^ n \ cdot e ^ x \ to + \ infty $ strives.
  • For $ x \ to - \ infty $, $ x ^ n \ cdot e ^ x \ to 0 $ strives, i.e. the x-axis is the asymptote of the graph of f with $ f (x) = x ^ n \ cdot e ^ x $.

example 1

$ f (x) = (x ^ 2-1) e ^ {- 2x} $

\ begin {align *}
\ lim_ {x \ to + \ infty} \ quad \ underbrace {(x ^ 2-1)} _ {\ rightarrow + \ infty} \ cdot \ underbrace {e ^ {- 2x}} _ {\ rightarrow 0} \ quad & \ rightarrow 0 \ \
\ lim_ {x \ to - \ infty} \ quad \ underbrace {(x ^ 2-1)} _ {\ rightarrow + \ infty} \ cdot \ underbrace {e ^ {- 2x}} _ {\ rightarrow + \ infty } \ quad & \ rightarrow + \ infty
\ end {align *}

Remember: When looking at the limit behavior, we orientate ourselves on the e-function - the most rapidly growing function.

Example 2

If we look at the graph of $ f (x) = (x ^ 2-1) e ^ {- 2x} $, our limit value calculation is confirmed.

  • if we let x run towards $ - \ infty $, the function tends towards + $ \ infty $
  • if we let x run towards $ \ infty $, the function tends towards 0, so the x-axis is asymptote

Daniel explains the limit behavior with an e-function again in his learning video.

Limit behavior for e-functions, Limes notation for e to the power of x | Math by Daniel Jung

Profile tasks with e-function

Think of the steps in profile tasks. It may be that the function you are looking for is the form

\ begin {align *}
f (x) = a \ cdot e ^ {- kx}
\ end {align *}

should have. There are therefore two unknowns and the task should have two conditions. In our example, the function should go through the points P (2 | 4) and Q (5 | 200). We thus set up our system of equations

\ begin {align *}
\ text {I} & \ quad \ quad 4 = a \ cdot e ^ {- 2k} \
\ text {II} & \ quad 200 = a \ cdot e ^ {- 5k}
\ end {align *}

and solve it for the unknowns a and k. Possibility: Change the equation $ \ text {I} $ to a and insert it in $ \ text {II} $. We then get for k = -1.3 and a = 0.6 and thus the function we are looking for:

\ begin {align *}
f (x) = 0.6 \ times e ^ {1.3 \ times x}
\ end {align *}

A simple example would be if the function you were looking for is the form

\ begin {align *}
f (x) = 4 \ cdot e ^ {- kx}
\ end {align *}

and through the point P (2 | 10) should. Why easier? Because there is only one unknown k.

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How to set up an e-function using 2 points, Daniel shows you here in his learning video.

Setting up an exponential function using 2 points, exponential function | Math by Daniel Jung

You can find more in-depth videos in Daniel's playlist on the subject of e-function!

Playlist: e-function, the special exponential function, Euler function, analysis