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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I had understood that 7 ln 7 = ln 6^7 And that ln 6 - ln 2 = ln 6/2 (or 3)

But if you have:

7 ln 6 - ln 3

and you first write it as 7 ln 6/3 you have:

7 ln 2

If you first rewrite it as

ln 6^7 - ln 3

then you have:

ln (6^7)/3

But this produces a different answer then 7 ln 2

This seems to indicate you should use the power/coefficent rule last, but what if you had:

5 ln 3 - 4 ln 2

you'd have to use the power coefficiant rule first.

I think I'm just forgetting a few things. I haven't done logarithmic equations in while.

A logarithm is just a misspelled algorithm.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

You need to use the power-coefficient rule first.

The flaw in your reasoning is that you have assumed that 7 ln 2 was the correct answer, but it is not. 7 ln 2 would be the answer if the question was 7(ln 6 - ln 3).

The simpler equivalent would be 7(6) - 3. Obviously, you'd need to make that into 42 - 3 = 39, but the flawed way of doing it would be to make it 7(6-3) = 7(2) = 14.

Why did the vector cross the road?

It wanted to be normal.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Huh? I assumed 7 ln 2 was the correct answer because that was the answer in my book! Hang on let me check it again.

In the meantime heres another problem I can't solve:

(e^x - e^-x)/2 = 7

e^x -e^-2 = 14

e^x = 14 + e^-x

ln e^x = ln 14 + ln e^-x

x ln e = ln 14 - x ln e

x = ln 14 -x (ln e = 1)

2x = ln 14

x = (ln 14 )/2

Does this look right to you?

In my book the answer is ln (7 + sqrt(50))

That doesn't make any sense. I've never found an incorrect answer in my mathbook before, let alone two in one lesson.

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Heres the first problem:

gold is being mined at a rate of R(t) = 7/t ounces per hour. How much gold is removed between the third hour and the six hour of operation?

The gold removed equals rate of removal multiplied by the time spent working. Therefore, its the area under the curve of 7/t at the intervals 3 and 6.

So lets integrate. 7/t : 7 ln t

Area = 7 ln 6 - 7 ln 3

= 7 (ln 6 - ln 3)

Yeah I must have forgotten the 7 on one of those.

Still, what about the one I just mentioned? (e^x - e^-x)/2 = 7 ?

*Last edited by mikau (2005-12-13 13:18:10)*

A logarithm is just a misspelled algorithm.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

(e^x - e^-x)/2 = 7

My calculator gives the same answer as your book, but I'm darned to heck if I know how it got there.

El que pega primero pega dos veces.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Not only that, but evaluation of that answer does not seem to produce 7. Perfect! X_X

*Last edited by mikau (2005-12-13 18:42:01)*

A logarithm is just a misspelled algorithm.

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**Flowers4Carlos****Member**- Registered: 2005-08-25
- Posts: 106

hi yaz mikau!!!

i'm sorry but i'm having trouble solving (e^x - e^-x)/2 = 7

also i can't seem to find a value for x to make that statement true!!

however, you did make one mistake:

e^x = 14 + e^-x

ln e^x = ln 14 + ln e^-x

you cannot distribute the log because that would imply: ln 14 + ln e^-x ⇒ 14e^-x which is not true.

the correct way of applying logs would be:

e^x = 14 + e^-x

ln e^x = ln (14 + e^-x)

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**Flowers4Carlos****Member**- Registered: 2005-08-25
- Posts: 106

hooray!!! i solved it!!!

(e^x - e^-x)/2 = 7

e^x - e^-x = 14 multiply everything by e^x

e^x(e^x - e^-x) = 14e^x

e^2x - 1 = 14e^x

(e^x)² - 14e^x - 1 = 0 let w=e^x

w² - 14w - 1 = 0 using the quadratic formula, i obtain

14±(200)^1/2)

w= ------------------ which simplifies to

2

w= 7±(50)^(1/2) plug w back in w=e^x

7±(50)^(1/2) = e^x

ln(7±(50)^(1/2)) = x but lnx > 0

ln(7+(50)^(1/2)) = x

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Aww.. dang you got here before me. I solved it this morning over a cup of coffee. (relaxation defined) Yeah the key is to multiply the equation by e^x and write it in the form of a quadratic, au^2 + bu + c where u = e^x. The solution to the quadratic will be 7 +- sqrt(50), it can't be 7 - sqrt (50) because that would be a negative number and e^x can never be negative. Thus 7 + sqrt(50) is the only solution.

Oh and yes it is the solution. I was checking it incorrectly.

What still bugs me is attempting to solve it with logarithms did not work. Disturbing.

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

But I'm not sure I see the mistake I made which you pointed out.

e^x = e^-x + 14

Oh... your saying I have to take the ln of both sides, not the ln of each term. Just like squaring or taking the square root of both sides.

e^x = ln (e^-x + 14)

Now I remember. Thanks for that, man!

A logarithm is just a misspelled algorithm.

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