multi variables and ln

JohnS

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Trying to solve for t:
t=d/(C*(M*(-0.072*ln(t)+1.2745)))

I took a first step to:
t/d=1/(C*(M*(-0.072*ln(t)+1.2745)))

2. d/t=C*(M*(-0.072*ln(t)+1.2745))

3. d= (C*(M*(-0.072*ln(t)+1.2745)))/t

And I'm stuck

Tried running this through some online calculators with no luck. Closest, I got was: t=e^(W-1(125d/(9Ce^(2549/144)*M))+2549/144)
I don't even understand "W". Here is the link to the calculator: http://calculator.mathcaptain.com/solving-equations-with-variables-on-both-sides-calculator.html

Thanks in advance.
 
Trying to solve for t:
t=d/(C*(M*(-0.072*ln(t)+1.2745)))

I took a first step to:
t/d=1/(C*(M*(-0.072*ln(t)+1.2745)))

2. d/t=C*(M*(-0.072*ln(t)+1.2745))

3. d= (C*(M*(-0.072*ln(t)+1.2745)))/t

And I'm stuck

Tried running this through some online calculators with no luck. Closest, I got was: t=e^(W-1(125d/(9Ce^(2549/144)*M))+2549/144)
I don't even understand "W". Here is the link to the calculator: http://calculator.mathcaptain.com/solving-equations-with-variables-on-both-sides-calculator.html

Thanks in advance.

Try a simpler example: \(\displaystyle t = ln(t)\)

Can't be done.

The real question is, can you get something useful out of it even after you know that? Mr. Lambert did some of the work for you back in 1700s. You tripped over it a little. https://en.wikipedia.org/wiki/Lambert_W_function
 
Try a simpler example: \(\displaystyle t = ln(t)\)

Can't be done.

The real question is, can you get something useful out of it even after you know that? Mr. Lambert did some of the work for you back in 1700s. You tripped over it a little. https://en.wikipedia.org/wiki/Lambert_W_function

Thank you. I had posted hoping for a simple answer, but instead got introduced to a whole new mathematical concept I never knew about. Much better in my book.

Cheers!
 
This is another example of no direct solution (a financial formula):

f = a*r / [(1+r)^n - 1] : solve for r
 
Try a simpler example: \(\displaystyle t = ln(t)\)

Can't be done.

The real question is, can you get something useful out of it even after you know that? Mr. Lambert did some of the work for you back in 1700s. You tripped over it a little. https://en.wikipedia.org/wiki/Lambert_W_function

So, this problem made me realize that I was looking at things wrong in the first place. I was trying to hit a moving target from a moving platform. I reworked the formula and this is what I have, but I get very confused about cancelling and extracting data from logarithms.

F=((-0.072*LN(t)+1.2745)^(3.25*(-0.072*LN(t)+1.2745))/60*t*100)*(W)*(1.5*(((d*1000)/(t*60))/(7.9917*d^-0.103)))/(C/((((((0.2*((d*1000)/t)+3.5)/3.5)*M)/60)*17.48)/((-0.072*LN(t)+1.2745))))^2

Still trying to solve for t.
 
F=((-0.072*LN(t)+1.2745)^(3.25*(-0.072*LN(t)+1.2745))/60*t*100)*(W)*(1.5*(((d*1000)/(t*60))/(7.9917*d^-0.103)))/(C/((((((0.2*((d*1000)/t)+3.5)/3.5)*M)/60)*17.48)/((-0.072*LN(t)+1.2745))))^2

Still trying to solve for t.
Are you trying to use the LambertW function? Is that what symbol W is for? What is F?

I asked a computer algebra system (MVR5), to solve the original equation for t. It gives a single power of e as the solution (with output from the LambertW[] function as part of the exponent). I don't know how to achieve that solution by hand. It defines LambertW as the function which satifies the following relationship.

LambertW[x] * e^LambertW[x] = x

MVR5 says there is no solution for your latest equation (treating symbols W and F as numbers). I also asked Wolfram-Alpha to solve it. Wolfram can interpret the equation by itself, but gets confused trying to understand a request to solve it for t.
 
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Are you trying to use the LambertW function? Is that what symbol W is for? What is F?

I asked a computer algebra system (MVR5), to solve the original equation for t. It gives a single power of e as the solution (with output from the LambertW[] function as part of the exponent). I don't know how to achieve that solution by hand. It defines LambertW as the function which satifies the following relationship.

LambertW[x] * e^(LambertW[x]) = x

MVR5 says there is no solution for your latest equation (treating symbols W and F as numbers). I also asked Wolfram-Alpha to solve it. Wolfram can interpret the equation by itself, but gets confused trying to understand a request to solve it for t.

Sorry for the ambiguity. That is not a Lambert W. I should have replaced it with a b. F is also a number. I think the calculators get hung up with the ln(t) in the exponent. When I put in values, I get an answer for F, so shouldn’t I be able to reverse it and get an answer for t if everything else is constant? Or is that not a safe assumption?
 
When I put in values, I get an answer for F, so shouldn’t I be able to reverse it and get an answer for t if everything else is constant?
NO. Not directly.
Once more, try and solve this one for r:
f = a*r / [(1+r)^n - 1]

If a=1000, r=.02, n=12, then f=~74.56

Now solve for r; you'll see it can't be done directly.

Not convinced? Try it using Wolfram;)
 
NO. Not directly.
Once more, try and solve this one for r:
f = a*r / [(1+r)^n - 1]

If a=1000, r=.02, n=12, then f=~74.56

Now solve for r; you'll see it can't be done directly.

Not convinced? Try it using Wolfram;)

Stupid math. Why can't it just be easy?

You say, "can't be done directly" does that imply that there is an indirect way?

Edit: Checked it on Wolfram and got an answer for your equation. It came up with [invalid links removed]

So, that tells me that it can be done, but I still don't understand how. And Wolfram still doesn't come up with a solution on my equation.
 
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Stupid math. Why can't it just be easy?

You say, "can't be done directly" does that imply that there is an indirect way?

"Directly" is not quite the right word.

The fact is that most questions you can ask in math can't be answered exactly. We teach the equations that can be solved by algebra; we tend to skip over the many more equations that can't, like these. This sort can only be solve numerically -- that is, by approximation. Schools are starting to realize this, and teach things that require technology or other methods that are used to solve real-life problems, rather than just the "pet" problems that we have tamed. But too many people still think math can do anything.
 
Stupid math. Why can't it just be easy?

You say, "can't be done directly" does that imply that there is an indirect way?
Yes. A great many problems cannot be solved directly through arithmetic, algebra, calculus, etc, but can be solved approximately through what are called numerical methods, which is a special course usually taught in college primarily to math majors. With my academic field having been political and economic history and my practical experience having been in finance, I found numerical methods to be one of my most helpful math courses.

People are primarily taught how to solve the types of problems that can be solved directly, whether exactly or approximately. These types of problems have many practical applications, but it is unfortunate that people are not taught that many other types of practical problems exist and that many of those can be solved approximately using indirect methods.
 
… Why can't it just be easy?
This must be a rhetorical question (like asking, "Why can't everything in the universe be perfect?") ;)

You're trying to do something that can't be done. You're looking for an exact solution that does not exist. The math is not stupid; it's logical. It provides multiple ways to form approximations for t, to whatever precision you'd like.


You say, "can't be done directly" does that imply that there is an indirect way?
Denis meant there's no way to find an exact expression for t. The indirect way is to use approximation methods. That's what our software did, when it gave us that solution in terms of the LambertW function. You now have an expression to approximate t.

Lambert.JPG


Edit: Checked it on Wolfram and got an answer for [Denis' financial] equation. It came up with [invalid URL link] …
I'm not sure what you entered, but Denis' equation has no exact solution.

WAeqn.JPG WAeqnFail.JPG


We can see that Wolfram interprets the equation, but it cannot provide an algebraic solution for r.
 
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