If 3^(2n) = (1/9)^(n+2), what is the value of n?

3^(2n) = 3^(-2n - 4)
Equate the powers: 2n = -2n - 4
4n = -4
n = -1

Answer: B

1/9= 3^-2

So, 3^(2n) = 3^(-2n - 4)
Equating the powers: 2n = -2n - 4
4n = -4
n = -1

B is the answer

Using the negative exponent rule, we have:

3^2n = 9^(-n - 2)

3^2n = 3^(-2n - 4)

With a common base of 3 for both expressions, we can equate the exponents:

2n = -2n - 4

4n = -4

n = -1

Answer: B

Since the answer options represent the value of n, back-solving can be a great strategy to solve this question.

But before starting with the options, let us simplify the equation given to ensure that we have the same bases on both sides.

\(3^{2n}\) can be written as \((3^2)^n\) i.e. \(9^n\)
.
\((\frac{1}{9})^{n+2}\) can be written as\( \frac{1 }{ 9^{n+2}}\) since 1 raised to any power is always 1.

Therefore, the equation can be rewritten as, \(9^n\) =\(\frac{ 1 }{ 9^{n+2}}\)

Let’s start with answer option C i.e. n = 0. If n = 0, LHS = \(9^0\) = 1 and RHS = \(\frac{1 }{ 9^3}\).

Clearly, both are not equal since the LHS is an integer and the RHS is a fraction. Answer option C can be eliminated. Further, it helps us understand that we cannot use any positive value for n since it will create a similar situation.
This helps us eliminate answer options D and E. We are left with answer options A and B.

If n = -1, LHS = \(9^{-1}\) = \(\frac{1}{9}\) and RHS =\( \frac{1 }{ 9^{1}}\) = \(\frac{1}{9}\). LHS = RHS, so the equation is satisfied.

Answer option A can be eliminated since there can only be one right answer.
The correct answer option is B.

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