How do you deal with the different bases when solving the equation:
$$\log_3 (2 - 3x) = \log_9 (6x^2 - 19x + 2)$$
I'm going round in circles trying to reconcile the bases.
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2Hint: $a=b\Leftrightarrow 9^a=9^b$. – Tomas Jul 15 '13 at 02:21
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2Hint: $\log_3 a = \frac{1}{2}\log_9 a$. – Thomas Andrews Jul 15 '13 at 02:22
2 Answers
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Raise both sides to the $9$th power to get:
$$ 9^{\log_3(2 - 3x)} = 9^{\log_9 (6x^2 - 19x + 2)} $$
$$ (3^2)^{\log_3(2 - 3x)} = 6x^2 - 19x + 2 $$ $$ (3^{\log_3(2 - 3x)})^2 = 6x^2 - 19x + 2 $$ $$ (2-3x)^2 = 6x^2 - 19x + 2$$
This is a quadratic in $x$. Solve appropriately.
One last thing (thanks to Calvin Lin): Keep in mind that in order for the $\log x$ to be defined (over the real numbers), then $x > 0$. After you solve your quadratic, you want to check for extraneous roots. Make sure that your solutions for $x$ satisfy that $ 2 - 3x > 0$ and $ 6x^2 - 19x + 2 > 0 $. (You shouldn't have any problems here, but it's important to do so in general!)
George V. Williams
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1Should also mention that you need to check the domain, i.e. $2 - 3x > 0$ and $6x^2-19x+2>0$ – Calvin Lin Jul 15 '13 at 02:37
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HINT:
Let $$\log_3 (2 - 3x) = \log_9 (6x^2 - 19x + 2)=a$$
$\implies 9^a=6x^2 - 19x + 2$ and $3^a=2-3x\implies 9^a=(3^2)^a=(3^a)^2=(2-3x)^2$
So, we have $6x^2 - 19x + 2=(2-3x)^2$
which is a Quadratic Equation of $x$ on re-arrangement
lab bhattacharjee
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