Solve
4^y -7(2^y) = 8
4^y = (2^2)^y = (2^y)^2
so, we have a quadratic in 2^y:
(2^y)^2 - 7*2^y - 8 = 0
(2^y-8)(2^y+1) = 0
y=3
2^y is never negative, so 2^y+1=0 has no real solution.
To solve the equation 4^y - 7(2^y) = 8, we can use algebraic techniques.
Step 1: Let's simplify the equation by rearranging the terms:
4^y - 7(2^y) - 8 = 0
Step 2: Notice that both terms involve different bases (4 and 2), so it's helpful to convert them to the same base. Since 4 = 2^2, we can rewrite 4^y as (2^2)^y, which simplifies to 2^(2y):
2^(2y) - 7(2^y) - 8 = 0
Step 3: Now, let's make a substitution to simplify the equation further. Let's introduce a new variable u = 2^y. This substitution transforms the equation into:
u^2 - 7u - 8 = 0
Step 4: This is now a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. In this case, let's factor the quadratic:
(u - 8)(u + 1) = 0
Step 5: Now, we can solve for u by setting each factor equal to zero and solving the resulting linear equations:
u - 8 = 0 or u + 1 = 0
Solving the first equation, u - 8 = 0, we get u = 8.
Solving the second equation, u + 1 = 0, we get u = -1.
Step 6: Recall that we defined u = 2^y. We can now solve for y by substituting the values of u back into the equation:
When u = 8, 2^y = 8. We can rewrite 8 as 2^3, giving us 2^y = 2^3. Since the bases (2) are the same, their exponents must be equal. Therefore, y = 3.
When u = -1, we can't find a real value of y since 2^y is always positive and -1 is negative.
Step 7: Therefore, the only solution to the equation 4^y - 7(2^y) = 8 is y = 3.