# Logarithmic Equation Help

• Feb 15th 2008, 01:57 PM
funnylookinkid
Logarithmic Equation Help
http://img.photobucket.com/albums/09...899e5f9b21.png

Howdy friends --

I'm in business calculus at a college in New Jersey and I recently was given this problem to solve on an online homework assignment. The question is to solve for x in terms of k and there is only one line for an answer.

I have already tried submitting x=9^k and x=(9^k)-10, and neither are satisfactory to complete the problem.

Many thanks,
flk
• Feb 15th 2008, 02:47 PM
Jhevon
Quote:

Originally Posted by funnylookinkid
http://img.photobucket.com/albums/09...899e5f9b21.png

Howdy friends --

I'm in business calculus at a college in New Jersey and I recently was given this problem to solve on an online homework assignment. The question is to solve for x in terms of k and there is only one line for an answer.

I have already tried submitting x=9^k and x=(9^k)-10, and neither are satisfactory to complete the problem.

Many thanks,
flk

remember your formulas for summing logs, and taking the inverse of logarithms.

$\log_9 x + \log_9 (x + 10) = k$

$\Rightarrow \log_9 [x(x + 10)] = k$

$\Rightarrow 9^k = x(x + 10)$

• Feb 15th 2008, 04:04 PM
funnylookinkid
Thank you so much! It worked brilliantly. :) I never would have thought to set the equation up that way!
• Feb 15th 2008, 04:32 PM
Krizalid
Quote:

Originally Posted by Jhevon
$\Rightarrow 9^k = x(x + 10)$

Well from here we can save some of time by just completing the missing square.

$9^k+25=x^2+10x+25\implies9^k+25=(x+5)^2.$ The rest follows.