Question

9^n+2 : 81^n * 3^n+3 = 1

Answer 1

1) 81^n • 3^n + 3 = 3^4n • 3^n + 3 = 3^5n + 3
2) 9^n + 2 = 3 ^2n + 4
3) ( 3^2n + 4 ) : ( 3^5n + 3 ) = 3 ^ ( 2n + 4 - 5n - 3 ) = 3 ^ ( - 3n + 1 )
4) 1 = 3 ^ 0
5) 3 ^ ( - 3n + 1 ) = 3 ^ 0
- 3n + 1 = 0
3n = 1
n = 1/3

Answer 2

I assume you want to solve the equation:

(9^n + 2) / (81^n * 3^(n+3)) = 1

To solve for n, we can simplify the left-hand side by using the properties of exponents:

(9^n + 2) / (81^n * 3^(n+3)) = [(9/81)^n * 9^2] / [(3^3)^n * 3^n] = (1/9)^n * 9^2 / 3^(3n) * 3^n = (1/9)^n * 3^-2

Now we can set this equal to 1 and solve for n:

(1/9)^n * 3^-2 = 1

(1/9)^n = 3^2

nlog(1/9) = 2log(3)

n = 2*log(3) / log(1/9)

Using the change of base formula for logarithms, we can rewrite this as:

n = -2*log(3) / log(9)

So the solution is:

n = -2*log(3) / log(9) ≈ -0.4048

Note that this solution is negative, which means that the left-hand side of the equation is not defined for any positive value of n. Therefore, there is no solution in the set of positive real numbers.