Solve the equation

log₃ (2 – 3x) = log₉ (6x^2 – 19x + 2)

How can I solve this? Please help.

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- Jan 5th 2008, 12:58 AMgetonProblem in exponential and logarithms
Solve the equation

log₃ (2 – 3x) = log₉ (6x^2 – 19x + 2)

How can I solve this? Please help. - Jan 5th 2008, 01:21 AMmr fantastic
The log bases haven't formatted properly.

*Assuming*the same base on each side (if they're not, then you're gonna have to make it much clearer what the bases are) it follows that:

2 - 3x = 6x^2 - 19x + 2

=> 6x^2 - 16x = 0

=> .......

But be careful - check all potential solutions by subbing into the original equation. One of them must be discarded ...... - Jan 5th 2008, 01:50 AMwingless
$\displaystyle \log _ab = \log _{\left(a^2\right)}\left(b^2\right)= \log _{\left(a^3\right)}\left(b^3\right)= \log _{\left(a^n\right)}\left(b^n\right)$

Using this, $\displaystyle \log _3(2-3x) = \log _9\left(4-12x+9x^2\right)$

Then,

$\displaystyle \log _9\left(4-12x+9x^2\right) = \log _9\left(6x^2-19x+2\right)$

$\displaystyle 4-12x+9x^2 = 6x^2-19x+2$

Now you can solve this equation for x.

(Don't forget that $\displaystyle 4 - 12 x + 9 x^2 > 0, 6x^2-19x+2 > 0)$ - Jan 5th 2008, 02:30 AMwgunther

an easy way to do this is to imagine both in an exponent of base 9. so

$\displaystyle 9^{\log_3(2-3x)}=9^{\log_9(6x^2-19x+2)}$

$\displaystyle (3^2)^{\log_3(2-3x)}=9^{\log_9(6x^2-19x+2)}$

$\displaystyle 3^{2\cdot \log_3(2-3x)}=9^{\log_9(6x^2-19x+2)}$

$\displaystyle 3^{ \log_3((2-3x)^2)}=9^{\log_9(6x^2-19x+2)}$

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

Now just solve, and check the solutions actually make sense (don't take logs of negative numbers, etc)