Logarithms Practice Test 2

Question : 18 of 39

Marks: +1, -0

The equation

\log_e x + \log_e (1 + x) = 0

can be written as:

Solution:

\log_e x + \log_e (1 + x) = 0

⇒ \log_e x(1 + x) = 0

⇒x(1+x)=e^0=1

⇒x^2+x-1=0

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