Question

Solve: log2(x-1)+log2(x+5)=4

Answer 1

Using the properties of logarithms, we can simplify the left-hand side of the equation:

log2(x-1) + log2(x+5) = log2((x-1)(x+5))

Therefore, the equation becomes:

log2((x-1)(x+5)) = 4

Using the definition of logarithms, we can rewrite this equation as:

2^4 = (x-1)(x+5)

16 = x^2 + 4x - 5

Simplifying further:

x^2 + 4x - 21 = 0

We can now use the quadratic formula to solve for x:

x = (-4 ± sqrt(4^2 - 4(1)(-21))) / (2*1)

x = (-4 ± sqrt(100)) / 2

x = (-4 ± 10) / 2

x = -7 or x = 3

However, we need to check if these solutions satisfy the original equation.

When x = -7:

log2(x-1) + log2(x+5) = log2((-7-1)(-7+5)) = log2(16) = 4

So x = -7 is a valid solution.

When x = 3:

log2(x-1) + log2(x+5) = log2((3-1)(3+5)) = log2(16) = 4

So x = 3 is also a valid solution.

Therefore, the solutions to the equation log2(x-1) + log2(x+5) = 4 are x = -7 and x = 3.

Explanation:

Answer 2

Using the logarithmic identity log(a) + log(b) = log(ab), we can simplify the left-hand side of the equation:

log2(x-1) + log2(x+5) = log2((x-1)(x+5))

So the equation becomes:

log2((x-1)(x+5)) = 4

Using the exponential form of logarithms, we can rewrite the equation as:

2^4 = (x-1)(x+5)

Simplifying:

16 = x^2 + 4x - 5

Rearranging:

x^2 + 4x - 21 = 0

Using the quadratic formula:

x = (-4 ± sqrt(4^2 - 4(1)(-21))) / (2(1))

x = (-4 ± sqrt(100)) / 2

x = (-4 ± 10) / 2

So x = -7 or x = 3.

However, we need to check whether these solutions satisfy the original equation. We can see that x = -7 does not work, because both terms inside the logarithms would be negative. Therefore, the only solution is x = 3.