Solve the equation
3^2x - 12 x 3^x + 27 = 0
let 3^x = y
so we get
y^2 - 12y + 27 = 0
(y-3)(y-9) = 0
y = 3 or y = 9
then
x^2 = 3 or x^2 = 9
x = ± √3 or x = ± 3
btw, in something like 12 x 3^x , do not use x as a multiplication sign, especially if you have a variable x
either say just 12(3^x) or say 12*3^x
To solve the equation 3^(2x) - 12x3^x + 27 = 0, we can use a substitution.
Let y = 3^x. The equation then becomes:
y^2 - 12xy + 27 = 0.
This is now a quadratic equation in terms of y. To solve for y, we can use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a).
In this case, a = 1, b = -12x, and c = 27. Plugging these values into the quadratic formula, we get:
y = (-(-12x) ± √((-12x)^2 - 4(1)(27))) / (2(1)).
y = (12x ± √(144x^2 - 108)) / 2.
y = 6x ± √(144x^2 - 108).
Now, we have two possible values for y. We can substitute these back into our original equation and solve for x.
To solve the equation 3^(2x) - 12x3^x + 27 = 0, we can use substitution and factoring.
Let's denote y as 3^x. Substituting this into the equation, we get:
y^2 - 12xy + 27 = 0
Now, we can factor this quadratic equation. We need to find two numbers that multiply to give 27 and add up to -12. These numbers are -3 and -9 since (-3) * (-9) = 27 and (-3) + (-9) = -12.
Therefore, the equation can be factored as follows:
(y - 3)(y - 9) = 0
Now we can set each factor equal to zero and solve for y:
y - 3 = 0 or y - 9 = 0
Solving each equation separately:
y = 3 or y = 9
Now we substitute back the value of y:
3^x = 3 or 3^x = 9
To solve 3^x = 3, we can rewrite it as an exponential equation with the same base:
3^(x - 1) = 1
Since any number raised to the power of 0 is equal to 1, we have:
x - 1 = 0
x = 1
To solve 3^x = 9, we can rewrite it as:
3^x = 3^2
Since the bases on both sides are the same, we can equate the exponents:
x = 2
Therefore, the solutions to the equation 3^(2x) - 12x3^x + 27 = 0 are x = 1 and x = 2.