Solve the equation
log₃ (2 – 3x) = log₉ (6x^2 – 19x + 2)
How can I solve this? Please help.
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 ......
$\displaystyle \log _ab = \log _{\left(a^2\right)}\left(b^2\right)= \log _{\left(a^3\right)}\left(b^3\right)= \log _{\left(a^n\right)}\left(b^n\right)$
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 $\displaystyle 4 - 12 x + 9 x^2 > 0, 6x^2-19x+2 > 0)$
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)