Formulae with logarithmic terms

Click For Summary

Homework Help Overview

The discussion revolves around solving an equation involving logarithmic terms, specifically the equation w = 1/h ln(l/lo - 1), with given values for w, lo, and h. Participants are attempting to find the variable L.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss substituting known values into the equation and manipulating logarithmic expressions. There are attempts to isolate L and check the validity of steps taken, with some questioning the correctness of their algebraic manipulations.

Discussion Status

Several participants have identified potential errors in their calculations and are revisiting their steps. There is an ongoing exploration of different interpretations of the logarithmic manipulation and the implications of assumptions made about the values involved. Some guidance has been offered regarding checking calculations and ensuring proper algebraic distribution.

Contextual Notes

Participants note the importance of ensuring that the arguments of logarithmic functions remain positive, leading to discussions about the constraints on L based on the values provided. There is also mention of the necessity to verify results by substituting back into the original equation.

brenfox
Messages
71
Reaction score
1

Homework Statement


w = 1/h ln(l/lo-1)
w=-2.6, lo=16 and h =1.5 Find L.

Homework Equations

The Attempt at a Solution


Plug in the values
2.6 = 1/1.5 ln(l/16 - 1)
Make the log term the subject.
ln(L/16 -1) = 1/1.5/-2.6
ln(L/16-1) = 0.666/-0.256
ln(L/16-1) = -0.256

Change the log statement to an index statement

L/16-1 = e^-0.256
Get L on its own
L = e^-0.256x16+1
L = 13.386
Am I on the right track?
 
Physics news on Phys.org
You can check by filling in the numbers, right?
First stumble is then that L/l0-1 < 0 and you can't take the ln of a negative number.
Conclusion: you're off the right track. Can you see where this last error popped up ?
Correct it and try again (check by filling in the numbers)
 
Forgot to plug in my answer! I will have another look at it. Seems like I am way off so will have to do a bit more reading into it. Thanks for the reply.
 
Yes. Just plugged in my answer and calculator shows error. So L must be bigger than 16.
 
L/16-1 = e^-0.256
Get L on its own
L = e^-0.256x16+1
Do this one step at a time

Then check - and if it still doesn't fit:

2.6 = 1/1.5 ln(l/16 - 1)
Make the log term the subject.
ln(L/16 -1) = 1/1.5/-2.6
Check this step.
 
L/16 -1 = e^-0.256

L -1 = e^-0.256 x 16

L = 1 e^-0.256 x 16

L = 1 + 0.774 x 16 = 28.3284. This is also wrong when i plug it in.
 
brenfox said:
ln(L/16 -1) = 1/1.5/-2.6

brenfox said:
L/16 -1 = e^-0.256

L -1 = e^-0.256 x 16

Is ##\frac{L}{16}-1=e^{-0.256}## EQUIVALENT TO ##L - 1=e^{-0.256} \times 16##?

Why didn't you multiply [1] by [16] ??
 
Last edited by a moderator:
brenfox said:
...

2.6 = 1/1.5 ln(l/16 - 1)
Make the log term the subject.
ln(L/16 -1) = 1/1.5/-2.6
...
To make the log term the subject, why not simply multiply (both sides) by 1.5 ?
 
I am starting to confuse myself here!

The equation is -2.6 = 1/1.5 ln(L/Lo -1)

multiply both sides by 1.5

-2.6x1.5 = ln(L/16 -1)

-3.9= ln(L/16-1)

L/16-1 e^-39

L/16 -1 = 0.02

L-1 =0.02 x 16

L = 0.02x16 +1
L= 1.32. I know i am going wrong somewhere because my answer does not plug into the original equation?
 
  • #10
brenfox said:
L/16 -1 = 0.02

L-1 =0.02 x 16

Again ,, Why don't you multiply 1 by 16 ?
You have to multiply 16 By each term in the equation ,, RIGHT ,,!
 
  • #11
If L-1 = 0.02 x 16 Then to get L on its own does the -1 turn into 1 on the other side of the equation? I can`t see the wood for the trees on this one.
 
  • #12
brenfox said:
If L-1 = 0.02 x 16 Then to get L on its own does the -1 turn into 1 on the other side of the equation? I can`t see the wood for the trees on this one.
-2.6x1.5 = ln(L/16 -1)

-3.9= ln(L/16-1)

L/16-1 = e^-3.9

L/16 -1 = 0.02 [Multiply EACH term by 16]

L-16 =0.02 x 16

L = 0.02x16 +16To justify the answer , plug in the value of L in the original equation ,

:)
 
Last edited:
  • #13
brenfox said:
I am starting to confuse myself here!

The equation is -2.6 = 1/1.5 ln(L/Lo -1)

multiply both sides by 1.5

-2.6x1.5 = ln(L/16 -1)

-3.9= ln(L/16-1)

L/16-1 e^-39
The step above is missing the =, and the step below is wrong. When you multiply both sides of the equation by 16, you need to multiply each term on the left side by 16. IOW, you have to distribute the 16 across both terms.
brenfox said:
L/16 -1 = 0.02

L-1 =0.02 x 16

L = 0.02x16 +1
L= 1.32. I know i am going wrong somewhere because my answer does not plug into the original equation?
 
  • #14
Thanks Maged, Mark. 16.32 plugs in correctly.
 
  • Like
Likes   Reactions: Maged Saeed

Similar threads

Calculate L using Omega, Lo = 16, and h = 1.5
  • · Replies 4 ·
Replies
4
Views
2K
Solving for time in an exponential equation
  • · Replies 12 ·
Replies
12
Views
2K
How Do You Calculate the Total Area of Rectangles and Triangles Combined?
  • · Replies 3 ·
Replies
3
Views
2K
Equation for the amplitude of a line sweeping across a triangle
  • · Replies 6 ·
Replies
6
Views
3K
Undergrad Logarithmic terms in a system of equations
  • · Replies 4 ·
Replies
4
Views
2K
Why Does \(\ln \frac{(x+1)}{(x-1)} \geq 0\) Imply \(x > 1\)?
  • · Replies 13 ·
Replies
13
Views
4K
How Do We Simplify and Interpret the Expression e^(ln(4)/2)?
  • · Replies 3 ·
Replies
3
Views
2K
Convergence of series log(1-1/n^2)
  • · Replies 2 ·
Replies
2
Views
4K
Pre-calc related question for the interval of a solution
  • · Replies 11 ·
Replies
11
Views
2K
Hydrogen Atom in an electric field along ##z##
  • · Replies 0 ·
Replies
0
Views
2K