# Problem in exponential and logarithms

• Jan 5th 2008, 01:58 AM
geton
Problem in exponential and logarithms
Solve the equation

log₃ (2 – 3x) = log₉ (6x^2 – 19x + 2)

• Jan 5th 2008, 02:21 AM
mr fantastic
Quote:

Originally Posted by geton
Solve the equation

log₃ (2 – 3x) = log₉ (6x^2 – 19x + 2)

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, 02:50 AM
wingless
$\log _ab = \log _{\left(a^2\right)}\left(b^2\right)= \log _{\left(a^3\right)}\left(b^3\right)= \log _{\left(a^n\right)}\left(b^n\right)$

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

Then,

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

$4-12x+9x^2 = 6x^2-19x+2$
Now you can solve this equation for x.
(Don't forget that $4 - 12 x + 9 x^2 > 0, 6x^2-19x+2 > 0)$
• Jan 5th 2008, 03:30 AM
wgunther
Quote:

Originally Posted by geton
Solve the equation

log₃ (2 – 3x) = log₉ (6x^2 – 19x + 2)

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