What is 1-log_{4}(m-1)-log_{4}(m+1) ?

The solution to 1-log_{4}(m-1)-log_{4}(m+1) is 1-log_{4}((m+1)(m-1))

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simplify 1-log_{4}(m-1)-log_{4}(m+1)

simplify

1 - lo g_{4} (m - 1) - lo g_{4} (m + 1)

1 - lo g_{4} ((m + 1) (m - 1))

Solution steps

1 - lo g_{4} (m - 1) - lo g_{4} (m + 1)

Apply log rule:

lo g_{a} (x) + lo g_{a} (y) = lo g_{a} (x y)

- lo g_{4} (m + 1) - lo g_{4} (m - 1) = lo g_{4} ((m + 1) (m - 1))

= 1 - lo g_{4} ((m + 1) (m - 1))

lo g_{2} (3) - 3 lo g_{2} (x) + \frac{1}{2} lo g_{2} (x + 1)

ln (\frac{x + y}{x - y})

2 [3 ln (x) - ln (x + 1) - ln (x - 1)]

- lo g_{b} (\frac{1}{2})

ln (x^{2} y^{5})

What is 1-log_{4}(m-1)-log_{4}(m+1) ?
The solution to 1-log_{4}(m-1)-log_{4}(m+1) is 1-log_{4}((m+1)(m-1))