To solve the equation, we can start by rewriting the equation in exponential form:
3^(Log3(x^2-9x+47)) = 3^3
This simplifies to:
x^2 - 9x + 47 = 27
Now we have a quadratic equation, so we can set it equal to zero and solve for x:
x^2 - 9x + 47 - 27 = 0x^2 - 9x + 20 = 0
Now we can factor this quadratic equation:
(x - 4)(x - 5) = 0
Setting each factor equal to zero:
x - 4 = 0 or x - 5 = 0x = 4 or x = 5
Therefore, the solutions to the equation are x = 4 and x = 5.
To solve the equation, we can start by rewriting the equation in exponential form:
3^(Log3(x^2-9x+47)) = 3^3
This simplifies to:
x^2 - 9x + 47 = 27
Now we have a quadratic equation, so we can set it equal to zero and solve for x:
x^2 - 9x + 47 - 27 = 0
x^2 - 9x + 20 = 0
Now we can factor this quadratic equation:
(x - 4)(x - 5) = 0
Setting each factor equal to zero:
x - 4 = 0 or x - 5 = 0
x = 4 or x = 5
Therefore, the solutions to the equation are x = 4 and x = 5.