# Thread: logs in terms of x & y

1. ## logs in terms of x & y

Log (p^8q^6) =

Log p^-6/q^-2 =

Log p^3/Log q^7 =

(Log p^9)^-9 =

expand as much as possible, then replace each log with x and y.

i need help! puhhleeez

2. you haven't posted your enitre question, what are x and y defined as, I presume we have something along the lines of $x = \log p$ and $y = log q$

I'll help you with the first one to get you started. As these question are very routine.

$\log (p^8 q^6) = \log(p^8) + \log(q^6)$ using the multiplication law of logs.

then use the power law

$\log(p^8) + \log(q^6) = 8 \log(p) + 6 \log(6)$

the replace he logs with x and y (assuming I guessed their definitions correctly)

$\log (p^8 q^6) = 8x + 6y$

Now you need to use the power law and division law again for the next question.

Bobak

edit: everything you need is given in that PDF earboth attached in this thread http://www.mathhelpforum.com/math-he...829-post5.html

3. Originally Posted by bharriga
Log (p^8q^6) =

Log p^-6/q^-2 =

Log p^3/Log q^7 =

(Log p^9)^-9 =

expand as much as possible, then replace each log with x and y.

i need help! puhhleeez
$
\log(p^8q^6) = \log p^8 + \log q^6$
$\leftarrow$ I have used the product-sum rule $\log(ab) = \log a + \log b$

$\log p^8 + \log q^6 = 8\log p + 6\log q$ $\leftarrow$ I have used the exponents rule $\log(a^k) = k\log a$

Try the remaining questions. They are all similar. First bring the terms inside the log to product form to apply the product-sum rule.Then, use the exponent rule to get the exponents out. Then try to simplify if possible.

If you get stuck somewhere, post your problem here, we will help you

By the way, what x and y are you talking about here:
expand as much as possible, then replace each log with x and y.