Question

help please!
solve for y: 3^y+1= 9^y-1
1. 1
2. 2
3. 3
4. 4

Answer 1

hi :DDD

Answer 2

You have written \(3^{y} + 1 = 9^{y} - 1\) You will have grave difficulty solving this. Please remember your Order of Operations and use parentheses to write what you mean.

\(3^{y+1} = 9^{y-1}\)

\(3^{y+1} = \left(3^{2}\right)^{y-1}\)

\(3^{y+1} = 3^{2y-2}\)

\(y+1 = 2y-2\)