9^x+2 - 6* 3^x+1= 0 find x
If you let u=3^x then the equation becomes (using parentheses to clarify matters)
9^(x+2) - 6*3^(x+1) = 0
9^2*9^x - 6*3*3^x = 0
81u^2-18u+1=0
(9u-1)^2 = 0
u = 1/9
So, 3^x = 1/9 = 3^-2
x = -2
Actually how do you get +1 after81u^2-18u +1 I can't understand
To solve the equation 9^(x+2) - 6 * 3^(x+1) = 0 for x, let's break down the solution step-by-step:
Step 1: Simplify the equation as much as possible.
We can rewrite 9^(x+2) as (3^2)^(x+2) = 3^(2(x+2)) = 3^(2x+4).
Similarly, 3^(x+1) can be written as 3^x * 3^1 = 3^x * 3.
Our equation now becomes: 3^(2x+4) - 6 * 3^x * 3 = 0.
Step 2: Combine like terms.
We can combine the two terms with 3^x by factoring out 3^x:
3^x * (3^2 - 6 * 3) = 0.
Step 3: Simplify further.
3^x * (9 - 18) = 0.
Step 4: Evaluate the inside parentheses.
3^x * (-9) = 0.
Step 5: Divide both sides by -9 to isolate 3^x.
3^x = 0.
Step 6: Solve for x.
There is no x that can satisfy 3^x = 0 because any number raised to the power of 0 is always equal to 1.
Therefore, there is no solution for the given equation.