Problem in exponential and logarithms

**geton** · January 5th 2008, 12:58 AM

Solve the equation

log₃ (2 – 3x) = log₉ (6x^2 – 19x + 2)

How can I solve this? Please help.

**mr fantastic** · January 5th 2008, 01:21 AM

Originally Posted by geton

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 ......

**wingless** · January 5th 2008, 01:50 AM

$\log _ab = \log _{\left(a^2\right)}\left(b^2\right)= \log _{\left(a^3\right)}\left(b^3\right)= \log _{\left(a^n\right)}\left(b^n\right)$

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 $4 - 12 x + 9 x^2 > 0, 6x^2-19x+2 > 0)$

**wgunther** · January 5th 2008, 02:30 AM

Originally Posted by geton

Solve the equation

log₃ (2 – 3x) = log₉ (6x^2 – 19x + 2)

How can I solve this? Please help.

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

$9^{\log_3(2-3x)}=9^{\log_9(6x^2-19x+2)}$
$(3^2)^{\log_3(2-3x)}=9^{\log_9(6x^2-19x+2)}$
$3^{2\cdot \log_3(2-3x)}=9^{\log_9(6x^2-19x+2)}$
$3^{ \log_3((2-3x)^2)}=9^{\log_9(6x^2-19x+2)}$
$(2-3x)^2=6x^2-19x+2$
Now just solve, and check the solutions actually make sense (don't take logs of negative numbers, etc)

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