# Algebra: Laws of Logs

1. Aug 2, 2009

### CentreShifter

I need to solve for t and it's slipped my mind how to manipulate this.

$$\frac{1}{4}=te^{-8t}$$

to

$$ln(1/4)=lnt-8t$$

I understand the laws of logs (I think), but I still can't seem to isolate the t.

2. Aug 2, 2009

3. Aug 2, 2009

### jgens

Well, you could also prove that there isn't any value of t that satifies that equation. It's pretty simple using derivatives . . .

Proof: Consider the function f defined such that f(t) = exp(8t) - 4t. Clearly, if for some value t0 we have that f(t0) = 0 then 4-1 = t0exp(-8t0). Now, we need only prove that there is no value t0 with this property.

From the definition of f(t) we certainly have that f '(t) = 8exp(8t) - 4. Since f '(t) is defined for all real numbers, the only critical points can occur when f '(t0) = 8exp(8t0) - 4 = 0 => t0 = - ln(2)/8. Since f '(t) < f '(t0 for t < t0 and f '(t) > f '(t0 for t > t0, we know that f(t0) is an absolute minimum. However, because f(- ln(2)/8) = 2-1(1 + ln(2)) > 0, there is no value t0 such that 4-1 = t0exp(-8t0). Q.E.D.

4. Aug 3, 2009

### John Creighto

I think I can express the derivative in terms of the inverse function in terms of itself f(y)=t

$$f(te^{-8t})=t$$

Differentiating both sides:

$$f'(te^{-8t})\left(e^{-8t}-8te^{-8t}\right)=1$$

$$f'(te^{-8t})=\frac{1}{e^{-8t}-8te^{-8t}}$$

$$f'(y)=\frac{1}{\frac{1}{f(y)}y-8y}$$

Which might have some potential for numeric algorithms or evaluating the power series.

Lets try a big more. At the point you are are interested in y=1/4

$$f'(1/4)=\frac{1}{\frac{1}{t}(1/4)-8(1/4)}$$

$$f'(1/4)=\frac{1}{\frac{1}{t}(1/4)-2}$$

$$\frac{1}{t}(1/4)-2=\frac{1}{f'(1/4)}$$

$$t=\frac{1}{4}\frac{1}{\frac{1}{f'(1/4)}+2}$$

Not sure if this helps at all.

Last edited: Aug 3, 2009
5. Aug 3, 2009

6. Aug 3, 2009

### Gib Z

That's somewhat hard to read, I'm not sure what exactly you did.

I was thinking of something like : f'(x) = 0 at x=1/8 only , f''( 1/8) < 0, hence f( 1/8) is a max. Evaluating, it is clearly less than 1/4.

7. Aug 3, 2009

### jgens

Well, here's a revised edition of the proof. I'm new to LaTeX so it doesn't look very neat . . .

Statement: $$(\forall t)(t \in \mathbb {R})\left (\frac{t}{e^{8t}} \neq \frac{1}{4} \right )$$

Proof: Consider the function $$f$$ defined such that $$f(t) = e^{8t} - 4t$$. From the definition of $$f$$, if for some $$t = t_{0}$$ we have that $$f(t_{0}) = 0$$, then

$$\frac{t_{0}}{e^{8t_{0}}} = \frac{1}{4}$$​

Now, consider the first derivative of $$f$$ so that $$f '(t) = 8e^{8t} - 4$$. Since $$f '$$ is defined $$\forall t \in \mathbb {R}$$, the only critical points can occur when $$f '(t_{0}) = 0$$. Solving for the critical points, we find that $$e^{8t_{0}} = 1/2$$. Since $$e^{8t} > 0$$, we don't lose any solutions when taking the logarithm base $$e$$ of the previous functions, therefore $$t_{0} = - ln(2)/8$$. Using a similar method, we can show that $$f '(t) < 0$$ if $$t < t_0$$ and similarly $$f '(t) > 0$$ if $$t > t_0$$. This proves that $$f(t_0)$$ is an absolute minimum. However,

$$f(t_0) = e^{-ln(2)} + \frac{ln(2)}{2} = \frac{1}{2} + \frac{ln(2)}{2} = \frac{1}{2}(1 + ln(2)) > 0$$​

which proves that there is no value of $$t$$ such that $$f(t) = 0$$.

So, I don't think that there's any need to invoke the lambert W function or power series, just really simple calculus. :)

8. Aug 3, 2009

### g_edgar

Complex solutions exist. The formula is not elementary, but can be written using the Lambert W function. t = -W(-2)/8 . One value is about -0.02160200035 - 0.2092108018 i

9. Aug 4, 2009

### John Creighto

It should be $$f(t) = e^{-8t} - 4t$$. which pretty much invalidates the rest of your post.

10. Aug 4, 2009

### John Creighto

Cool, thanks. :) I worked this out on wikipedia and got the same thing.