# Thread: Logarithms Solving for x

1. ## Logarithms Solving for x

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

2. Originally Posted by DaCoo911
i have a problem here that has stumped me. any help is greatly appreciated, thanks in ADVANCE!

Use the rules $log_b(a) + log_b(c) = log_b(ac)$ and $klog_b(a) = log_b(a^k)$

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

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

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

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

3. O NO i'm sorry! these two are not added together they are subtracted. WHich means there is division, not multiplication. SORRY!

4. Originally Posted by DaCoo911
O NO i'm sorry! these two are not added together they are subtracted. WHich means there is division, not multiplication. SORRY!
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!