2+log√1+x+3log√1-x=log √1-x^2. Solve for x?

1 Answer
Apr 11, 2018

x=0.99

Explanation:

I'm assuming you wish for us to solve

2+log1+x+3log1x=log1x2

We must apply the logarithm laws:

2+log1+x+3log1x=log(1+x)(1x)

2+12log(1+x)+32log(1x)=12log((1+x)(1x))

2+12log(1+x)+32log(1x)=12log(1+x)+12log(1x)

log(1x)=2

102=1x

x=1102

x=11100

x=99100=0.99

Hopefully this helps!