# Log problems

• Nov 30th 2012, 11:11 AM
agehayoshina
having squared logs problems
The question states that log3q = p

i have to write the answer in the form of P

-1/2logq(81q²)

thanks!
• Nov 30th 2012, 11:29 AM
becauseifeelikeit
Re: Log problems
We can write -1/2 logq⁡(81q^2 )=p as logq[(81q^2 )^-1/2], based on the laws of logs. This can be simplified to logq(1/9q) = P. This can be further simplified to logq(1/9) + logq(1/q). We know that logq(1/q) is simply -1, so we can rewrite the equation as logq(1/9) - 1 = P.Bringing 1 over to the other side gives logq(1/9) = P + 1. We can use the change of base formula to write logq(1/9) as log(1/9)/logq, giving the equation log(1/9)/logq = P + 1. Now we multiply both sides by logq to clear the equation of the denominator, resulting in log(1/9) = (P + 1)logq. We divide by (P + 1) in order to isolate q, giving logq = log(1/9)/(P + 1). Now we simply write the equation in exponential form, and find that 10^[log(1/9)/(P + 1)] = q.

Hope this helps!
• Nov 30th 2012, 03:57 PM
Soroban
Re: having squared logs problems
Hello, agehayoshina!

I hope I understand the problem.
Here is what I think it says . . .

Quote:

$\text{Given: }\,\log_3(q) = p$
$\text{write in terms of }p\!:\;-\tfrac{1}{2}\log_q(81q^2)$

We have: . $-\tfrac{1}{2}\log_q(81q^2) \;=\;-\log_q(81q^2)^{\frac{1}{2}} \;=\;-\log_q(9q)$

. . . . . . $=\;-\big[\log_q(9) + \log_q(q)\big] \;=\;-\big[\log_q(9) + 1\big] \;=\;-\log_q(9) - 1$

. . . . . . $=\;-\frac{\log_3(9)}{\log_3(q)} - 1 \;=\; -\frac{2}{p} - 1$