What is 2-1/2 log_{n}(4)-log_{n}(5) ?

The solution to 2-1/2 log_{n}(4)-log_{n}(5) is 2-log_{n}(10)

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simplify 2-1/2 log_{n}(4)-log_{n}(5)

simplify

2 - \frac{1}{2} lo g_{n} (4) - lo g_{n} (5)

2 - lo g_{n} (10)

Solution steps

2 - \frac{1}{2} lo g_{n} (4) - lo g_{n} (5)

Apply log rule:

b \cdot lo g_{a} (x) = lo g_{a} (x^{b})

- \frac{1}{2} lo g_{n} (4) = - lo g_{n} (4^{\frac{1}{2}})

= 2 - lo g_{n} (4^{\frac{1}{2}}) - lo g_{n} (5)

Simplify

- lo g_{n} (4^{\frac{1}{2}}) - lo g_{n} (5) : - lo g_{n} (10)

= 2 - lo g_{n} (10)

2 ln (5) + \frac{7}{5} - 2 ln (- 3) - \frac{7}{3}

5 lo g_{10} (a x) + 3 lo g_{10} (a) y

\frac{19}{2} lo g_{2} (\frac{7}{5}) - \frac{19}{2}

lo g_{r} (11 x)

What is 2-1/2 log_{n}(4)-log_{n}(5) ?
The solution to 2-1/2 log_{n}(4)-log_{n}(5) is 2-log_{n}(10)