# Simultaneous equations help

1. Feb 14, 2006

### discombobulated

I need some help with this please!

1. given p=logq 16, express in terms of p:
a) logq 2
b) logq (8q)

this is what i got:
a) p= logq16 = logq22
= 4logqqq
=1/4 p

b) p=logq16 =logq8 +logq q
=logq23 + 1
=3logq2 +1
....??

Also can someone please help get me started on these similtaneous equations?

8y =42x+3
log2 y = log2x +4

I'm not sure exactly how to equate them

2. Feb 14, 2006

### Kamataat

For the first one, the answer ($\log_q2=p/4$) is OK, but I have trouble understanding what you wrote before that.

For the second one, you can't assume that 16=8q. Otherwise, you're on the right track:

$$\log_q(8q)=\log_q8+\log_qq=3\log_q2+1$$

You already know what $\log_q2$ is in terms of p, so you're done.

EDIT:

For the third one, try to isolate y in terms of x from the second equation. Remember that $\log_216=4$. Then substitute that expression into the first equation and rewrite the eq. as a logarithm in base 8. Solve for x and then y.

EDIT2: Tnx, arildno, no wonder I didn't understand what he wrote, since it's wrong.

- Kamataat

Last edited: Feb 15, 2006
3. Feb 15, 2006

### VietDao29

I'm not quote sure what you are doing here. But since, 8 = 23, and 4 = 22. So taking the logarithm base 2 of both sides, gives:
8y = 42x + 3
<=> log2(8y) = log2(42x + 3)
<=> log2(23y) = log2(22(2x + 3))
You can go from here, right? :)

4. Feb 15, 2006

### arildno

a) is totally wrong.
You are in effect saying p=1/4p, which is incorrect.

5. Feb 16, 2006

### discombobulated

ok i'm sorry for not setting it out clearly enough. What i meant was:
log 9 2 = 1/4 p
which according to my text book is correct

don't worry i've managed to work out b) as well
if log 9 2 = 1/4 p
then logq(8q) = logq (23q)
so logq(8q) = 3/4p +1

Last edited: Feb 16, 2006
6. Feb 16, 2006

### VietDao29

Have you worked out the equation one yet?
Is it clear, or do you need more hints? :)