If log_2(x) + log_2(x-1) = 3, what is the value of x?

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Q: If log_2(x) + log_2(x-1) = 3, what is the value of x?

Solution: log_2(x(x-1)) = 3 => x(x-1) = 2^3 = 8 => x^2 - x - 8 = 0. Solving gives x = 5.

Steps: 12

Step 1: Start with the equation log_2(x) + log_2(x-1) = 3.
Step 2: Use the property of logarithms that says log_a(b) + log_a(c) = log_a(b*c). So, rewrite the equation as log_2(x(x-1)) = 3.
Step 3: Convert the logarithmic equation to an exponential form. This means x(x-1) = 2^3.
Step 4: Calculate 2^3, which equals 8. So now we have x(x-1) = 8.
Step 5: Expand the left side: x^2 - x = 8.
Step 6: Rearrange the equation to set it to zero: x^2 - x - 8 = 0.
Step 7: Factor the quadratic equation. We need two numbers that multiply to -8 and add to -1. These numbers are -4 and 2.
Step 8: Write the factored form: (x - 4)(x + 2) = 0.
Step 9: Set each factor to zero: x - 4 = 0 or x + 2 = 0.
Step 10: Solve for x. From x - 4 = 0, we get x = 4. From x + 2 = 0, we get x = -2.
Step 11: Since x must be greater than 1 (because of the log), we discard x = -2.
Step 12: Therefore, the solution is x = 4.