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If 32x+1−3x=3x+3−323^{2x + 1} - 3^x = 3^{x + 3} - 3^232x+1−3x=3x+3−32 then the values of xxx are:​
Question

If 32x+13x=3x+3323^{2x + 1} - 3^x = 3^{x + 3} - 3^2 then the values of xx are:​

A.

1, -2

B.

-1, -2

C.

0, 1

D.

2, -1

Correct option is D

Given:
32x+13x=3x+393^{2x + 1} - 3^x = 3^{x + 3} - 9​​
Concept used:
Use exponential laws and substitution to reduce the equation.
Rewrite powers
32x+1=3×(3x)23x+3=27×3x3^{2x + 1} = 3 \times (3^x)^2 \\3^{x + 3} = 27 \times 3^x​​
Let y=3x, y = 3^x,​ then the equation becomes:
3y2y=27y93y^2 - y = 27y - 9​​
Rearranging terms
3y2y27y+9=03y228y+9=03y^2 - y - 27y + 9 = 0 \\3y^2 - 28y + 9 = 0​​
Solve quadratic
Using the quadratic formula:
y=(28±sqrt(2824×3×9))/(2×3)=(28±sqrt(784108))/6=(28±(676))/6=(28±26)/6y = (28 ± sqrt(28^2 - 4 \times 3 \times 9)) / (2 \times 3) \\= (28 ± sqrt(784 - 108)) / 6 \\= (28 ± \sqrt(676)) / 6 \\= (28 ± 26) / 6​​
So,
y=54/6=9=>3x=9=>x=2y=2/6=1/3=>3x=1/3=>x=1y = 54 / 6 = 9 => 3^x = 9 => x = 2 \\y = 2 / 6 = 1/3 => 3^x = 1/3 => x = -1​​
Correct answer is (D: 2, -1

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