Solve for n:
((3^n)^n*(3^n)^-1) / 81 = 1
Answer (n) = -1, 4
I have no idea how n = -1, 4. Also, please don't ask if the answers are actually wrong or not, they are directly from the text book.
((3^n)^n*(3^n)^-1) / 81 = 1
3^(2n) * 3^-n = 81
3^n = 81
3^n = 3^4
n = 4
testing the n = -1 solution
Left Side
= ( (3^-1)^(-1) * (3^-1)^-1 )/81
= (3^1 * 3^1 )/81
= 9/81
≠ 1
So according to the way you typed it, the book is wrong,
the only solution is n = 4
Alright, thanks so much for the help!
To solve for n in the equation ((3^n)^n*(3^n)^-1) / 81 = 1, we can follow these steps:
Step 1: Simplify the equation.
First, let's simplify the expression ((3^n)^n * (3^n)^-1). Using the property of exponents (a^m * a^n = a^(m+n)), we can simplify this to (3^(n*n + n*-1)) = (3^(n^2 - n)).
Then, we have ((3^(n^2 - n)) / 81 = 1).
Step 2: Multiply both sides by 81.
By multiplying both sides of the equation by 81, we eliminate the fraction and get:
3^(n^2 - n) = 81.
Step 3: Rewrite 81 as 3^4.
Since both 81 and 3 are powers of 3, we can rewrite 81 as 3^4.
3^(n^2 - n) = 3^4.
Step 4: Set the exponents equal to each other.
In order for two exponential expressions to be equal, their exponents must be equal. So, we can set the exponents equal to each other:
n^2 - n = 4.
Step 5: Rearrange the equation and solve for n.
To solve the quadratic equation n^2 - n - 4 = 0, we can use factoring or the quadratic formula. Here, we'll use factoring:
n^2 - n - 4 = 0.
Factoring the equation, we can write it as:
(n - 2)(n + 2) = 0.
Setting each factor equal to zero, we get:
n - 2 = 0, or n + 2 = 0.
Solving for n in each case, we find:
n = 2, or n = -2.
Therefore, the solutions to the equation ((3^n)^n*(3^n)^-1) / 81 = 1 are n = 2 and n = -2.
However, it seems that the textbook provided the answers n = -1 and n = 4. It is possible that there was an error in the solution you provided or in the textbook answer key. It is recommended to double-check the calculations and the original problem to ensure the accuracy of the answers.