2+log√1+x+3log√1-x=log √1-x^2. Solve for x?
1 Answer
Explanation:
I'm assuming you wish for us to solve
2+log√1+x+3log√1−x=log√1−x2
We must apply the logarithm laws:
2+log√1+x+3log√1−x=log√(1+x)(1−x)
2+12log(1+x)+32log(1−x)=12log((1+x)(1−x))
2+12log(1+x)+32log(1−x)=12log(1+x)+12log(1−x)
log(1−x)=−2
10−2=1−x
x=1−10−2
x=1−1100
x=99100=0.99
Hopefully this helps!