Math problem in senior one: it is known that the function f(x)=log (1-x)+log (x+3) has a base (0 is less than a and less than 1).

f(x)=loga[( 1-x)(x+3)]

First, the domain name is -3.

1, f(x)=0, namely loga[( 1-x)(x+3)]=0.

Because 0 = loga (1);

So: loga [(1-x) (x+3)] = loga (1)

That is (1-x) (x+3) =1;

Finishing: x? +2x-2 = 0； Get: x 1=- 1-√3, x2 =-1+√ 3;

Both satisfy the domain -3

So the zero point of f(x) is-1√ 3;

2、f(x)=loga(-x？ -2x+3)

The symmetry axis of the real number is x=- 1, so the real number increases at (-3,-1); Decreasing upward (-1,1);

Because 0

So f(x) decreases at (-3,-1); Increment to (-1,1);

So: f (x) min = f (-1) = loga (4) =-4.

So: a (-4) = 4.

Get: a=√2/2

I hope I can help you. If you don't understand, please hi me and wish you progress in your study!