(log3 x)^2 -4 log3 x+3=0
To solve the equation (log3 x)^2 - 4 log3 x + 3 = 0, let's make a substitution.
Let's assume log3 x = y, such that x = 3^y.
Now, substitute this value back into the equation:
(y)^2 - 4(y) + 3 = 0.
To solve this quadratic equation, we can factor it:
(y - 1)(y - 3) = 0.
Setting each factor equal to zero gives us:
y - 1 = 0 or y - 3 = 0.
For y - 1 = 0,
y = 1.
Substituting y = 1 back into our assumption, we get:
log3 x = 1.
Now, let's solve for x:
x = 3^(1).
x = 3.
For y - 3 = 0,
y = 3.
Substituting y = 3 back into our assumption, we get:
log3 x = 3.
Now, let's solve for x:
x = 3^(3).
x = 27.
So the solutions to the equation are x = 3 and x = 27.
To solve the equation (log3 x)^2 - 4 log3 x + 3 = 0, we can let u = log3 x. This substitution will simplify the equation.
So, we have u^2 - 4u + 3 = 0.
To factorize the quadratic equation, we need to find two numbers whose product is 3 and whose sum is -4.
The numbers -1 and -3 satisfy these conditions, so we can rewrite the equation as (u - 1)(u - 3) = 0.
Now, we can apply the zero product property, which states that if the product of two factors is equal to zero, then at least one of the factors must be equal to zero.
So, we have two possible cases:
Case 1: u - 1 = 0.
Solving this equation, we find u = 1.
Case 2: u - 3 = 0.
Solving this equation, we find u = 3.
Now, we substitute back u = log3 x into the two cases:
Case 1: log3 x = 1.
To rewrite this equation in exponential form, we have 3^1 = x. Therefore, x = 3.
Case 2: log3 x = 3.
To rewrite this equation in exponential form, we have 3^3 = x. Therefore, x = 27.
So the solutions to the original equation are x = 3 and x = 27.