The equation loga(x) + loga(1+x)=0 can be written as ?

Step 1: Use the product rule for logarithms: log_a x + log_a(1+x) = log_a(x(1+x)).

Step 2: Set the combined logarithm equal to 0: log_a(x(1+x)) = 0. Since log_a 1 = 0, this is equivalent to log_a(x(1+x)) = log_a 1.

Step 3: Remove the logarithm (assuming base a>0 and a≠1 and arguments positive): x(1+x) = 1 → x^2 + x - 1 = 0.

• Solve the quadratic: x = [-1 ± √(1 + 4)]/2 = (-1 ± √5)/2.

• Domain check: logarithms require x>0 and 1+x>0, so x>0. Therefore only the positive root x = (-1 + √5)/2 is valid.

Conclusion: The expression log_a x + log_a(1+x) can be written as log_a 1 when set equal to 0, and solving gives x = (-1 + √5)/2 as the valid solution.