solve for x: 9^2x * 27^(3-x)=1/9?
2009-07-05 3:03 pm
Solve for x. I tend to keep getting lost when I am working on this equation.
回答 (4)
2009-07-05 3:09 pm
✔ 最佳答案
Change every base into a power of 3.9 becomes 3^2
27 is 3^3
1/9 = 1/3^2 = 3^(-2)
so now we have:
(3^2)^(2x) * (3^3)^(3-x) = 3^(-2)
Power to a power: multiply the powers
3^(4x) * 3^(9-3x) = 3^(-2)
Multiply by adding the powers:
3^(9+x) = 3^(-2)
Set powers equal:
9+x = -2
x = -11
2009-07-05 3:49 pm
9^2x * 27^(3-x)=1/9
3^2(2x) * 3^3(3-x)=3^-2
4x * 9-3x=-2
x+9=-2
x=-2-9
x=-11 answer//
3^2(2x) * 3^3(3-x)=3^-2
4x * 9-3x=-2
x+9=-2
x=-2-9
x=-11 answer//
2009-07-05 3:28 pm
9^(2x) * 27^(3 - x) = 1/9
(3^2)^(2x) * (3^3)^(3 - x) = 9^-1
3^(4x) * 3^[3(3 - x)] = (3^2)^-1
3^(4x + 3*3 - 3*x) = 3^-2
4x + 9 - 3x = -2
4x - 3x = -2 - 9
x = -11
(3^2)^(2x) * (3^3)^(3 - x) = 9^-1
3^(4x) * 3^[3(3 - x)] = (3^2)^-1
3^(4x + 3*3 - 3*x) = 3^-2
4x + 9 - 3x = -2
4x - 3x = -2 - 9
x = -11
2009-07-05 3:08 pm
3^(x+9) = 1/9
Take the logarithm to the base 3 of both sides:
x+9 = -2
Subtract 9 from both sides:
x = -11
Now test that this solution is appropriate by substitution into the original equation:
Check the solution x = -11:
3^(x+9) => 3^(9-11) = 1/9 ~~ 0.111111
1/9 => 1/9 = 1/9 ~~ 0.111111
So the solution is correct.
Thus, the solution is:
x = -11
Take the logarithm to the base 3 of both sides:
x+9 = -2
Subtract 9 from both sides:
x = -11
Now test that this solution is appropriate by substitution into the original equation:
Check the solution x = -11:
3^(x+9) => 3^(9-11) = 1/9 ~~ 0.111111
1/9 => 1/9 = 1/9 ~~ 0.111111
So the solution is correct.
Thus, the solution is:
x = -11
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