Please help me to solve the following problems

Titicamara · 2006-07-14 17:43:03

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.

numen · 2006-07-14 22:49:04

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?

Titicamara · 2006-07-15 06:13:19

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.

numen · 2006-07-16 02:34:03

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%.

John E. Franklin · 2006-07-16 09:53:51

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)

John E. Franklin · 2006-07-16 09:57:06

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.

numen · 2006-07-16 10:10:42

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

Titicamara · 2006-07-16 20:35:13

Thank you guys. Much appreciated.

Math Is Fun Forum

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

Please help me to solve the following problems

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

Re: Please help me to solve the following problems

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

Re: Please help me to solve the following problems

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

Re: Please help me to solve the following problems

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

Re: Please help me to solve the following problems

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

Re: Please help me to solve the following problems

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

Re: Please help me to solve the following problems

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

Re: Please help me to solve the following problems

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