eh? You must be overthinking this one.
3log3x - log3x = 2
2log3x = 2
log3x = 1
x = 3^1 = 3
3log3x - log3x = 2
2log3x = 2
log3x = 1
x = 3^1 = 3
Using the quotient rule of logarithms, we can rewrite the equation as a single logarithm:
log3 (x^3 / x) = 2
Simplifying further:
log3 (x^2) = 2
Now, we can exponentiate both sides with base 3 to eliminate the logarithm:
3^2 = x^2
9 = x^2
Taking the square root of both sides:
x = ±√9
Therefore, the solutions to the equation are:
x = 3 or x = -3
So, we have:
3log3 x - log3 x = 2
log3 (x^3) - log3 x = 2
Now, using another logarithmic identity log a - log b = log (a/b), we can rewrite the equation as:
log3 (x^3 / x) = 2
Simplifying further, we have:
log3 (x^2) = 2
Now, we can rewrite this equation in exponential form:
3^2 = x^2
Simplifying, we get:
9 = x^2
Taking the square root of both sides, we obtain two possible solutions:
x = ±3
Therefore, the solutions to the equation 3log3 x - log3 x = 2 are x = 3 and x = -3.