Rearranging equation

  • Thread starter martine
  • Start date
  • #1
27
0

Main Question or Discussion Point

How do I calculate t from the following equation?

z=a+b(e^k*t -1)

Guess this should be right to start with:

t=ln(z/b)

but what to do with the a and -1? and how do I calculate ln(z/b)? as lnz/lnb? I've tried thinking (didn't work *smirk*) and trial and error with simple numers (didn't work either)

thanks a lot!
 

Answers and Replies

  • #2
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
17
First, am I correct in presuming that you mean:

[tex]
z = a + b (e^{kt} - 1)
[/tex]

? If so, then you need to group the exponent with parentheses; the correct way to write it is z=a+b(e^(k*t) -1).

(Order of operations says you do exponentiation before multiplication)


Anyways, this is a chance to use stuff you've learned previously. :smile: It looks like you want to use the fact:

[tex]
x = b^y \rightarrow \log_b x = y
[/tex]

right? However, note that the exponentiation has to be by itself on one side of the equation. Do you know how to do that?


how do I calculate ln(z/b)?
Compute z/b then take the log, just like the expression states.
 
  • #3
27
0
Yes, that equation of yours is what I meant. Should have read the faw on how to write equations in a forum.

I also understand the equation you wrote further down, but I guess I'm really to stupid to rearrange teh whole equation to t=....
 

Related Threads for: Rearranging equation

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
6
Views
3K
Top