3^2x - 12 * 3^x + 27
find the stupid x
sry forgot to add... that expression equals 0
begin by recognizing that 3^(2x) = (3^x)^2
Now you can set z = 3^x
and substitute
z^2 - 12z + 27 = 0
is a standard quadratic
(z - 3)(z - 9) = 0
(could use quadratic eqn to solve, would yield this result)
substituting back in for z, we have the two roots
3^x - 3 = 0 <== (a)
3^x - 9 = 0 <== (b)
Solve for x using natural logs:
(a) x*ln3 = ln3
x = 1
(b) x*ln3 = ln9
x = ln9/(ln3) = 2
Cool.
You can't find x unless the expression is set equal to something.
Comments
begin by recognizing that 3^(2x) = (3^x)^2
Now you can set z = 3^x
and substitute
z^2 - 12z + 27 = 0
is a standard quadratic
(z - 3)(z - 9) = 0
(could use quadratic eqn to solve, would yield this result)
substituting back in for z, we have the two roots
3^x - 3 = 0 <== (a)
3^x - 9 = 0 <== (b)
Solve for x using natural logs:
(a) x*ln3 = ln3
x = 1
(b) x*ln3 = ln9
x = ln9/(ln3) = 2
Cool.
You can't find x unless the expression is set equal to something.