How to Solve Simultaneous Equations with Logarithms

[discombobulated]

I need some help with this please!

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

this is what i got:
a) p= log_q16 = log_q2²
= 4log_qqq
=1/4 p

b) p=log_q16 =log_q8 +log_q q
=log_q2³ + 1
=3log_q2 +1
...??

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

8^y =4^2x+3
log₂ y = log₂x +4

I'm not sure exactly how to equate them

 

[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

 

[VietDao29]

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

8^y =4^2x+3
log₂ y = log₂x +4

I'm not sure exactly how to equate them

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:
8^y = 4^{2x + 3}
<=> log₂(8^y) = log₂(4^{2x + 3})
<=> log₂(2^3y) = log₂(2^{2(2x + 3)})
You can go from here, right? :)

 

[arildno]

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

 

[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 textbook is correct

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

 

[VietDao29]

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