# Thread: Complicated Logarithms

1. ## Complicated Logarithms

e^x = 5x - 3

Thank you.

2. Originally Posted by PacManFrogg
e^x = 5x - 3
Thank you.
Hello,

by your equation you calculate the intercepting points of graphs of an exponential function and a line. I've attached a diagram to show the situation.
I don't know a method to solve this equation algebraically. Instead I would use an iteration(?) (I mean that you have to construct a convergent sequence to obtain the solution). I'll demonstrate a method which I know under the name Newton's method:
1. Transform the equation into a function: f(x)=e^x-5x+3
2. Choose a x-value for a start. Here I would say: $\displaystyle x_0 = 1$
3. The next x-value is calculated by:

$\displaystyle x_{n+1}=x_n-{f(x_n)\over f'(x_n)}$ where f'(x) is the first derivation of f. With your problem it is: f'(x) = e^x-5

So you get:

$\displaystyle x_{1}=1-{{e^1-5+3}\over {e^1-5}}\approx 1.3147...$

Now plug in this value for $\displaystyle x_n$ and you'll get the next x-value which comes closer to the solution:
x_1 = 1.3147...
x_2 = 1.4323...
x_3 = 1.4653...
x_4 = 1.4687...
x_5 = 1.4688...
x_6 = is the same as x_5. That means you substract zero from the previous x-value which is only possible if f(x) = 0. That was intended. So you've found one solution.

Change the x_0-value to 2 and you'll get the second solution: x = 1.74375...

Greetings

EB

3. Originally Posted by earboth
Hello,

by your equation you calculate the intercepting points of graphs of an exponential function and a line. I've attached a diagram to show the situation.
I don't know a method to solve this equation algebraically. Instead I would use an iteration(?) (I mean that you have to construct a convergent serie to obtain the solution). I'll demonstrate a method which I know under the name Newton's method:
1. Transform the equation into a function: f(x)=e^x-5x+3
2. Choose a x-value for a start. Here I would say: $\displaystyle x_0 = 1$
3. The next x-value is calculated by:

$\displaystyle x_{n+1}=x_n-{f(x)\over f'(x)}$ where f'(x) is the first derivation of f. With your problem it is: f'(x) = e^x-5

So you get:

$\displaystyle x_{1}=1-{{e^1-5+3}\over {e^1-5}}\approx 1.3147...$

Now plug in this value for $\displaystyle x_n$ and you'll get the next x-value which comes closer to the solution:
x_1 = 1.3147...
x_2 = 1.4323...
x_3 = 1.4653...
x_4 = 1.4687...
x_5 = 1.4688...
x_6 = is the same as x_5. That means you substract zero from the previous x-value which is only possible if f(x) = 0. That was intended. So you've found one solution.

Change the x_0-value to 2 and you'll get the second solution: x = 1.74375...

Greetings

EB
There is another solution near x=1.75

RonL

4. Originally Posted by CaptainBlack
There is another solution near x=1.75
RonL
Hello,

you are right.
But as I've mentioned above it is x = 1.74375...

Greetings

EB

5. Originally Posted by earboth
Hello,

you are right.
But as I've mentioned above it is x = 1.74375...

Greetings

EB
Sorry I missed that.

The problem can also be solved using the Lambert W function.

RonL

6. Originally Posted by CaptainBlack
The problem can also be solved using the Lambert W function.
RonL

Hello,

and äääähem. Well I've heard the name of Heinrich Lambert. But I don't know his famous W-function (I believe to remeber that it deals somehow with complex arguments and differential equations(?)), and worse: I wouldn't be able to explain how to use.

Greetings

EB

7. Originally Posted by earboth
Hello,

and äääähem. Well I've heard the name of Heinrich Lambert. But I don't know his famous W-function (I believe to remeber that it deals somehow with complex arguments and differential equations(?)), and worse: I wouldn't be able to explain how to use.

Greetings

EB
The function, $\displaystyle xe^x$ is bijective for $\displaystyle x\geq 0$ therefore it has an inverse $\displaystyle W(x)$. That is how the lambert function is defined. (It happens to be non-elementary).

8. Originally Posted by ThePerfectHacker
The function, $\displaystyle xe^x$ is bijective for $\displaystyle x\geq 0$ therefore it has an inverse $\displaystyle W(x)$. That is how the lambert function is defined. (It happens to be non-elementary).
Unfortunately in this case the x and W(x) involved are not >0 (worse still the
x involved is not even >-1/e) so we are in a region where W is multi-valued
(as we should expect since there are two real roots to this problem).

My mention of W was a sort of joke, theoretically it can be used to write
the solution in closed form, practically one is better off using numerical
methods to solve the problem.

RonL