Solve the equations
i) log2(x + 2) + log2 x = 3.
ii) 25^x + 5^x+1 + 6 = 0.
i) To solve the equation log2(x + 2) + log2 x = 3, we will use the properties of logarithms.
First, we can rewrite the equation using the property: log(a) + log(b) = log(a * b):
log2(x + 2) + log2 x = log2((x + 2) * x)
Next, we use another property of logarithms: log(a^n) = n * log(a). Applying this property to our equation, we get:
log2((x + 2) * x) = log2(2^3)
From here, we can equate the expressions within the logarithms to solve for x:
(x + 2) * x = 2^3
Expanding the equation, we have:
x^2 + 2x = 8
Rearranging the equation, we get:
x^2 + 2x - 8 = 0
Now, we can solve this quadratic equation. There are different methods you can use, such as factoring, completing the square, or using the quadratic formula. Let's use factoring:
(x + 4)(x - 2) = 0
Setting each factor equal to zero, we get:
x + 4 = 0 or x - 2 = 0
Solving for x in each case:
x = -4 or x = 2
So, the solutions to the equation log2(x + 2) + log2 x = 3 are x = -4 and x = 2.
ii) The equation 25^x + 5^(x+1) + 6 = 0 involves exponential terms.
Let's simplify the equation first. We can rewrite 5^(x+1) as 5^x * 5^1, which is 5 * 5^x. Then our equation becomes:
25^x + 5 * 5^x + 6 = 0
Next, we can simplify further by recognizing that 25 is equal to 5^2:
(5^2)^x + 5 * 5^x + 6 = 0
5^(2x) + 5 * 5^x + 6 = 0
Now, let's replace 5^x with a variable, say let's set z = 5^x.
The equation becomes:
z^2 + 5z + 6 = 0
This is a quadratic equation. Factoring it, we have:
(z + 2)(z + 3) = 0
Setting each factor equal to zero:
z + 2 = 0 or z + 3 = 0
Solving for z in each case:
z = -2 or z = -3
Now, we can substitute our original variable back in:
5^x = -2 or 5^x = -3
However, exponentials can only be positive, so there are no real solutions for x in this case.
Therefore, the equation 25^x + 5^(x+1) + 6 = 0 has no real solutions.