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#1 2006-07-14 17:43:03

Titicamara
Member
Registered: 2006-07-14
Posts: 7

Please help me to solve the following problems

1. 25=10x
2. 40+10e^x=200
3. x^0.3=12
4. e^lnx=8

AND


5. Using the production function Q=K^0.8L^0.4 when Q=2000, state the change that will take place in K when L is increased by 1%.

Thanks guys/gals and most important me please guide me with the answer as well as the workings. Thank you.

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#2 2006-07-14 22:49:04

numen
Member
Registered: 2006-05-03
Posts: 115

Re: Please help me to solve the following problems

1. x=25/10=5/2. [divide by ten on both sides]

2. 10e^x=160 -> e^x=16 -> x=ln(16) [subtract 40, divide ten, logaritm both sides and ln(e)=1]

3. ln(x^0,3)=ln(12) -> ln(x)=ln(12)/0,3 -> x=e^(ln(12)/0,3)  [logaritm both, down with 0,3; solve for x]

4. ln(x)*ln(e)=ln(8) so ln(x)=ln(8), thus x=8. [logaritm on both sides, ln(e)=1]


5. Do you mean Q=K^(0,8)*L^(0,4) or Q=K^(0,8(L)^(0,4)) or something else?


Bang postponed. Not big enough. Reboot.

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#3 2006-07-15 06:13:19

Titicamara
Member
Registered: 2006-07-14
Posts: 7

Re: Please help me to solve the following problems

numen wrote:

5. Do you mean Q=K^(0,8)*L^(0,4) or Q=K^(0,8(L)^(0,4)) or something else?

The question in my book states Q=K^0.8L^0.4 when Q=2000 state the change that will take place in K when L is increased by 1%.

The answer given is if L increases 1%, then K decreases 0.5% but without any workings. I don't understand at all.

Btw thanks for your earlier answers.

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#4 2006-07-16 02:34:03

numen
Member
Registered: 2006-05-03
Posts: 115

Re: Please help me to solve the following problems

Hmm, ok I'll take it as my first suggestion then and lets see what I get.

We have that


The second equation says that L has increased 1% (1,01) and gamma is the change in K, which we should solve. Set them equal to get rid of Q:

Since

we can divide with K^0,8 and L^0,4 to get

So,

Which is approximately 0,99504... in other words a decrease by 0,5%.


Bang postponed. Not big enough. Reboot.

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#5 2006-07-16 09:53:51

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: Please help me to solve the following problems

I checked 1,2,3,and 4 and they are right!!
For problem #3), e^((ln12)/.3) can be reduced to 12^(3 1/3).
.3 is reciprocal of 3 1/3.
12 = e^ln12

Last edited by John E. Franklin (2006-07-16 09:54:24)


igloo myrtilles fourmis

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#6 2006-07-16 09:57:06

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: Please help me to solve the following problems

Also another way to think about exponents is this,
which is equivalent to above ways, but easier for me to remember.
Say base^exponent = answer.
Then I remember that  log(answer)/log(base) = exponent
It's no different, but I remember it better somehow.


igloo myrtilles fourmis

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#7 2006-07-16 10:10:42

numen
Member
Registered: 2006-05-03
Posts: 115

Re: Please help me to solve the following problems

Nice John, 12^(3 1/3) is a much better answer :]


Bang postponed. Not big enough. Reboot.

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#8 2006-07-16 20:35:13

Titicamara
Member
Registered: 2006-07-14
Posts: 7

Re: Please help me to solve the following problems

Thank you guys. Much appreciated.

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