How do I solve for t using the Laws of Logs in this algebraic equation?

[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.

 

[junglebeast]

It's not easy,

http://mathworld.wolfram.com/LambertW-Function.html

 

[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.

 

[John Creighto]

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.

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&#039;(te^{-8t})\left(e^{-8t}-8te^{-8t}\right)=1

f&#039;(te^{-8t})=\frac{1}{e^{-8t}-8te^{-8t}}

f&#039;(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&#039;(1/4)=\frac{1}{\frac{1}{t}(1/4)-8(1/4)}

f&#039;(1/4)=\frac{1}{\frac{1}{t}(1/4)-2}

\frac{1}{t}(1/4)-2=\frac{1}{f&#039;(1/4)}

t=\frac{1}{4}\frac{1}{\frac{1}{f&#039;(1/4)}+2}

Not sure if this helps at all.

 

[John Creighto]

I think you can get a power series solution. Do a numerical guess to get something close. Find the taylor series about that point. The derivative of an inverse function is given here:

http://en.wikipedia.org/wiki/Inverse_functions_and_differentiation#Additional_properties

and for higher order derivatives you can use the following generalization of the chain rule.

http://en.wikipedia.org/wiki/Faà_di_Bruno's_formula

 

[Gib Z]

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.

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.

 

[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 &#039;(t) = 8e^{8t} - 4. Since f &#039; is defined \forall t \in \mathbb {R}, the only critical points can occur when f &#039;(t_{0}) = 0. Solving for the critical points, we find that e^{8t_{0}} = 1/2. Since e^{8t} &gt; 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 &#039;(t) &lt; 0 if t &lt; t_0 and similarly f &#039;(t) &gt; 0 if t &gt; 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)) &gt; 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. :)

 

[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

 

[John Creighto]

Proof: Consider the function f defined such that f(t) = e^{8t} - 4t.

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

 

[John Creighto]

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

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