Solve log2(x-1) = 5-log2(x+3) for x.
base 2.
Log2(x-1) = 5 - Log2(x+3).
Log2(x-1) + Log2(x+3) = 5
Log2((x-1)(x+3)) = 5
Log2(x^2+3x-x-3) = 5
Log2(x^2+2x-3) = 5
x^2 + 2x -3 = 2^5 = 32
x^2 + 2x -3 -32 = 0
x^2 + 2x -35 = 0
(x-5)(x+7) = 0
x-5 = 0
X = 5.
x+7 = 0
X = -7.
Solution set: X = -7, X = 5.
To solve the given equation, we can use logarithmic properties and algebraic manipulation. Here's a step-by-step explanation of how to solve it:
1. Start by using the logarithmic property log_a(b) - log_a(c) = log_a(b/c). Apply this property to the equation:
log2(x-1) = 5 - log2(x+3)
log2(x-1) + log2(x+3) = 5
2. Next, combine the logarithms on the left side using the logarithmic property log_a(b) + log_a(c) = log_a(b * c):
log2[(x-1)(x+3)] = 5
3. Rewrite the equation in exponential form, using the definition that log_a(b) = c is equivalent to a^c = b:
2^5 = (x-1)(x+3)
4. Simplify the equation:
32 = x^2 + 2x - x - 3
5. Combine like terms on the right side:
32 = x^2 + x - 3
6. Move all terms to one side to set up a quadratic equation:
x^2 + x - 35 = 0
7. Solve the quadratic equation. You can use factoring, completing the square, or the quadratic formula. In this case, factoring is the simplest approach. Factor the quadratic equation into two binomial factors:
(x + 7)(x - 5) = 0
8. Set each factor equal to zero and solve for x:
x + 7 = 0 or x - 5 = 0
Solve for x:
x = -7 or x = 5
9. Check the solution(s) by substituting them back into the original equation. Since the log function is not defined for negative numbers or zero, we can discard x = -7 as an extraneous solution:
Checking for x = -7: log2(-7-1) = 5 - log2(-7+3)
log2(-8) = 5 - log2(-4)
The left side is undefined, so x = -7 is not a valid solution.
10. Therefore, the solution to the equation log2(x-1) = 5 - log2(x+3) is x = 5.