# Logarithim Question

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• Oct 16th 2008, 05:07 PM
Bladeowen
Logarithim Question
This is actually pretty urgent but for some reason I cant get into urgent Hw section. Been working on this for hours. And its for tomorrow.(Headbang)

Ok question goes...

Solve:

Log (base3) (2-3x) = log (base9) (6x² -19x +2)

the answer is apparently -2 and -1/3. I can get -1/3 from solving a quadratic equation at the, but annoyinglyI get +2 as the other answer.

So far I have :

(lg (2-3x))/lg3 = (lg(6x² -19x +2))/lg 9
(lg 9) (lg [2-3x]) = (lg 3) (log [6x² -19x +2])

Im guessing this is where i go wrong because this is where the actual answers stop making sense

9 (lg [2-3x]) = 3(lg [6x² -19x +2])
9(2-3x) = 3 (6x² -19x +2)
18-27x = 18x² - 57x +6
0 = 18x² - 30x -12
0 = (18x +6)(x-2)

x=2, -1/3

2 cannot be the answer beccause that would mean a negative log.

Pleaseeee Help, it may easily be something silly because im very tired. :(

Thanks,
Owen
• Oct 16th 2008, 07:45 PM
Jhevon
Quote:

Originally Posted by Bladeowen
This is actually pretty urgent but for some reason I cant get into urgent Hw section. Been working on this for hours. And its for tomorrow.(Headbang)

Ok question goes...

Solve:

Log (base3) (2-3x) = log (base9) (6x² -19x +2)

the answer is apparently -2 and -1/3. I can get -1/3 from solving a quadratic equation at the, but annoyinglyI get +2 as the other answer.

So far I have :

(lg (2-3x))/lg3 = (lg(6x² -19x +2))/lg 9
(lg 9) (lg [2-3x]) = (lg 3) (log [6x² -19x +2])

Im guessing this is where i go wrong because this is where the actual answers stop making sense

9 (lg [2-3x]) = 3(lg [6x² -19x +2])
9(2-3x) = 3 (6x² -19x +2)
18-27x = 18x² - 57x +6
0 = 18x² - 30x -12
0 = (18x +6)(x-2)

x=2, -1/3

2 cannot be the answer beccause that would mean a negative log.

Pleaseeee Help, it may easily be something silly because im very tired. :(

Thanks,
Owen

using the change of base formula was a good idea. here's a better one:

note that if $\log_a b = c$, then $\log_{\sqrt{a}} \sqrt{b} = c$. that is, $\log_a b = \log_{\sqrt{a}} \sqrt{b}$.

(try to prove that!)

with that in mind, note that your problem simplifies to:

$\log_3 (2 - 3x) = \log_3 \sqrt{6x^2 - 19x + 2}$

$\Rightarrow 2 - 3x = \sqrt{6x^2 - 19x + 2}$

$\Rightarrow (2 - 3x)^2 = 6x^2 - 19x + 2$

i suppose you can take it from here

(and, of course, be sure to test your answers in the original problem)
• Oct 16th 2008, 11:55 PM
Bladeowen
That is Brilliant

Thank you so much.