Log 3 (x-2)+log 3x =log3 8

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To solve the equation log3 (x-2) + log3x = log3 8, we can combine the two logarithms on the left side using the product rule of logarithms, which states that log a + log b = log (a*b).

So, log3 (x-2) + log3x = log3 (x-2)*x

This simplifies to:

log3 (x^2 - 2x) = log3 8

Now, we can rewrite the right side of the equation as a logarithm:

log3 8 = log3 2^3

Since 8 is equal to 2^3, we can rewrite the equation as:

log3 (x^2 - 2x) = log3 2^3

Now, we can set the expressions inside the logarithms equal to each other:

x^2 - 2x = 2^3

x^2 - 2x = 8

Now, we can solve this quadratic equation by setting it equal to zero:

x^2 - 2x - 8 = 0

Factoring this equation, we get:

(x - 4)(x + 2) = 0

Setting each factor equal to zero gives us two possible solutions:

x - 4 = 0, x = 4
x + 2 = 0, x = -2

Therefore, the two possible solutions for x are x = 4 and x = -2.