Formulae with logarithmic terms

In summary: Yes. Just plugged in my answer and calculator shows error. So L must be bigger than 16.L/16-1 = e^-0.256Get L on its ownL = e^-0.256x16+1Do this one step at a timeThen 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/-2.6Check this step.ln(L/16 -1) = e^-0.256L/16 -1 = e
  • #1
brenfox
71
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
  • #2
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)
 
  • #3
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.
 
  • #4
Yes. Just plugged in my answer and calculator shows error. So L must be bigger than 16.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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:
  • #8
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 ?
 
  • #9
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 Maged Saeed

1. What is a logarithm?

A logarithm is the inverse function of exponentiation. It is used to solve equations involving exponential terms, and is written as logb(x), where b is the base and x is the argument.

2. How are logarithms used in formulae?

Logarithms are often used in formulae to simplify complex expressions and make them easier to solve. They can also be used to convert exponential equations into linear equations, making them more manageable.

3. What is the difference between natural logarithms and common logarithms?

Natural logarithms, denoted by ln(x), have a base of e, which is approximately equal to 2.718. Common logarithms, denoted by log(x) or log10(x), have a base of 10. This means that ln(x) and log(x) give different numerical values for the same argument.

4. Can logarithms be negative?

Yes, logarithms can be negative. This occurs when the argument is between 0 and 1, as the base must be raised to a negative power to equal the argument. For example, log2(0.5) = -1.

5. How do you solve equations with logarithmic terms?

To solve equations with logarithmic terms, you can use the properties of logarithms, such as the power rule and product rule, to simplify the equation. Then, you can use algebraic methods to isolate the variable and solve for its value.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
4
Views
1K
  • Precalculus Mathematics Homework Help
Replies
12
Views
1K
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
4
Views
1K
  • Precalculus Mathematics Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
918
  • Introductory Physics Homework Help
Replies
2
Views
1K
Replies
1
Views
1K
Back
Top