If 3^2n = (1/9)^(n+2) Solving for exponents

[rockrock]

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

I understand that the first step is to rewrite the bases.

I re-wrote the right side of the equation to: (3^2)(-n+2). Which is the reciprocal, with the sign of the exponent changed.

However, the correct reformulation is: (3^-2)(n+2).

I am confused by how I am to know that the "sign change" is supposed to occur at the base, not on the exponent? They yield opposite results - the correct answer is -1. The wrong answer is what I chose, 1.

[adiagr]

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

I understand that the first step is to rewrite the bases.

I re-wrote the right side of the equation to: (3^2)(-n+2). Which is the reciprocal, with the sign of the exponent changed.

However, the correct reformulation is: (3^-2)(n+2).

I am confused by how I am to know that the "sign change" is supposed to occur at the base, not on the exponent? They yield opposite results - the correct answer is -1. The wrong answer is what I chose, 1.

Hi,

Just one point.


When you re-wrote the right side of the equation, you wrote
(3^2)(-n+2)

whereas it should be:
(3^2)(-n-2)



In the explanation, it is (3^-2)(n+2)

so basically exponent part is: (-1)(2)(n+2)


Aditya

[rockrock]

Ahh I see. So basically, the correct way for me to rewrite is is not merely to place a "negative sign" before the exponent, but instead to distribute (-1). Hence -n-2 is the negative version of the exponent.

That really helps! thanks.

[mschwrtz]

Thanks Aditya.

[DavidV891]

Hi

Is there a way to rewrite the basis on the lest to = 1/9?

[RonPurewal]

Hi

Is there a way to rewrite the basis on the lest to = 1/9?

assuming "lest" = left...

sure. you just need a substitution--in other words, you need to be able to express 3 in terms of 1/9.

starting from 1/9, if you take the square root and the reciprocal (in either order), you get 3. thus, 3 = (1/9)^(1/2).

if you substitute (1/9)^(1/2) for 3 in the left-hand part, you'll then have 1/9 as the base number on both sides. (after that point, there are still some non-trivial steps involving exponent rules.)