solve the equation log3
(2-3x) = log9 (6x^2-19x+2)
since 9=3^2, log3 = 2*log9 and we have, using base 9 logs,
2log(2-3x) = log(6x^2-19x+2)
(2-3x)^2 = 6x^2 - 19x + 2
3x^2 + 7x + 2 = 0
(3x+1)(x+2) = 0
x = -2 or -1/3
To solve the equation log3(2-3x) = log9(6x^2-19x+2), we can use the properties of logarithms to simplify the equation and then solve for x.
First, let's use the property of logarithms that states loga(b) = c is equivalent to a^c = b. Applying this property to the given equation, we have:
3^(log3(2-3x)) = 9^(log9(6x^2-19x+2))
Now, simplify the bases on both sides of the equation:
(2-3x) = (6x^2-19x+2)
Now we have a simple quadratic equation. Rearrange it into standard form:
6x^2 - 19x + 2 - (2 - 3x) = 0
Simplify further:
6x^2 - 19x + 2 - 2 + 3x = 0
Combine like terms:
6x^2 - 16x = 0
Factor out an x:
x(6x - 16) = 0
From here, we have two possibilities:
1) x = 0
2) 6x - 16 = 0
Solving 6x - 16 = 0 for x:
6x = 16
x = 16/6
x = 8/3
Therefore, the equation log3(2-3x) = log9(6x^2-19x+2) has two solutions: x = 0 and x = 8/3.