# 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 $\displaystyle x = \log p$ and $\displaystyle y = log q$

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

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

then use the power law

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

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

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

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

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

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.