i have a problem here that has stumped me. any help is greatly appreciated, thanks in ADVANCE!

http://i546.photobucket.com/albums/h...911/Help-3.jpg

Printable View

- Mar 4th 2009, 10:38 AMDaCoo911Logarithms Solving for x
i have a problem here that has stumped me. any help is greatly appreciated, thanks in ADVANCE!

http://i546.photobucket.com/albums/h...911/Help-3.jpg - Mar 4th 2009, 10:48 AMe^(i*pi)
Use the rules $\displaystyle log_b(a) + log_b(c) = log_b(ac)$ and $\displaystyle klog_b(a) = log_b(a^k)$

$\displaystyle log_3(x^2(x-2)) = 2$

$\displaystyle x^2(x-2) = x^3-2x^2 = 3^2$

$\displaystyle x^3-2x^2-9 = 0$

and solve for x bearing in mind only solutions in which x > 2 counts for you can't have the log of a non-positive number - Mar 4th 2009, 10:53 AMDaCoo911
O NO i'm sorry! these two are not added together they are subtracted. WHich means there is division, not multiplication. SORRY!

- Mar 4th 2009, 11:05 AMstapel
That does not, however, change the methods or processes. You still start by

**log rules**to simplify the left-hand side, and then use**the definition of logarithms**to convert the equation to its equivalent exponential form. Then you solve the equation (which, for subtraction, is a**rational equation**).

If you get stuck, please reply showing how far you have gotten. Thank you! :D