Question

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

Answer 1

Answer:

x² + x - 1 = 0

Step-by-step explanation:

Since we given that

loga (x) + loga (1+x) = 0

∵  Log(a×b) = Log(a) + Log(b)

∴  loga ((x)(1+x))=0

∵ According to the definition of logarithms at the base of b

Logb(a) = c    ⇔    b^c = a

So

a^(0)  = ((x)(1+x))

1 = x² + x                ∵ a^(0) = 1

By rearranging the terms we get

x² + x - 1 = 0

Thus

x² + x - 1 = 0 is equivalent to loga (x) + loga (1+x) = 0

Ans is  x² + x - 1 = 0