Question

Can somebody solve the equation (sqrt5) ^(y+1) = y^ {(2y-1) /3}?

Answer 1

To solve the equation (sqrt5)^(y+1) = y^((2y-1)/3), follow these steps: rewrite the expressions using fractional exponents, equate the two expressions, take the cube on both sides to eliminate the cube root, equate the exponents, and solve the resulting equation for y.

Step-by-step explanation:

To solve the equation (sqrt5)^(y+1) = y^((2y-1)/3), we need to isolate the variable y. Here are the steps:

1. Rewrite (sqrt5)^(y+1) as (5^(1/2))^(y+1), which can be simplified as 5^((y+1)/2).
2. Next, rewrite y^((2y-1)/3) as (y^(2y-1))^(1/3).
3. Now we can equate the two expressions: 5^((y+1)/2) = (y^(2y-1))^(1/3).
4. To simplify further, take the cube on both sides to eliminate the cube root: (5^((y+1)/2))^3 = ((y^(2y-1))^(1/3))^3.
5. This gives us 5^((3(y+1))/2) = y^(2y-1).
6. Since the base of both sides is the same (5), we can equate the exponents.
7. We have (3(y+1))/2 = 2y-1.
8. Solve the resulting equation for y by rearranging the terms and solving for y.