What is 2log_{4}(y-1)-log_{4}(y+3)-1 ?

The solution to 2log_{4}(y-1)-log_{4}(y+3)-1 is log_{2}(y-1)-log_{4}(y+3)-1

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simplify 2log_{4}(y-1)-log_{4}(y+3)-1

simplify

2 lo g_{4} (y - 1) - lo g_{4} (y + 3) - 1

lo g_{2} (y - 1) - lo g_{4} (y + 3) - 1

Solution steps

2 lo g_{4} (y - 1) - lo g_{4} (y + 3) - 1

Simplify

2 lo g_{4} (y - 1) : lo g_{2} (y - 1)

= lo g_{2} (y - 1) - lo g_{4} (y + 3) - 1

lo g_{10} (a^{6} b^{4})

3 ln (2) + 2 ln (x) - 12 ln (x + 4)

3 lo g_{2} (x) + lo g_{2} (y) - 5 lo g_{2} (z)

lo g_{5} (1000) + lo g_{4 a} (64 a^{3})

lo g_{3^{x}} (+ 2 lo g_{3^{y} - l o g_{3^{z}} (x)} (x))

What is 2log_{4}(y-1)-log_{4}(y+3)-1 ?
The solution to 2log_{4}(y-1)-log_{4}(y+3)-1 is log_{2}(y-1)-log_{4}(y+3)-1