What is 5log_{2}(k)-8log_{2}(m)+10log_{2}(n) ?

The solution to 5log_{2}(k)-8log_{2}(m)+10log_{2}(n) is log_{2}((k^5n^{10})/(m^8))

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simplify 5log_{2}(k)-8log_{2}(m)+10log_{2}(n)

simplify

5 lo g_{2} (k) - 8 lo g_{2} (m) + 10 lo g_{2} (n)

lo g_{2} (\frac{k ^{5} n ^{10}}{m ^{8}})

Solution steps

5 lo g_{2} (k) - 8 lo g_{2} (m) + 10 lo g_{2} (n)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

5 lo g_{2} (k) = lo g_{2} (k^{5})

= lo g_{2} (k^{5}) - 8 lo g_{2} (m) + 10 lo g_{2} (n)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

8 lo g_{2} (m) = lo g_{2} (m^{8})

= lo g_{2} (k^{5}) - lo g_{2} (m^{8}) + 10 lo g_{2} (n)

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{2} (k^{5}) - lo g_{2} (m^{8}) = lo g_{2} (\frac{k ^{5}}{m ^{8}})

= lo g_{2} (\frac{k ^{5}}{m ^{8}}) + 10 lo g_{2} (n)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

10 lo g_{2} (n) = lo g_{2} (n^{10})

= lo g_{2} (\frac{k ^{5}}{m ^{8}}) + lo g_{2} (n^{10})

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{2} (\frac{k ^{5}}{m ^{8}}) + lo g_{2} (n^{10}) = lo g_{2} (n^{10} \frac{k ^{5}}{m ^{8}})

= lo g_{2} (n^{10} \frac{k ^{5}}{m ^{8}})

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= lo g_{2} (\frac{k ^{5} n ^{10}}{m ^{8}})

3 ln (A) - \frac{1}{2} [ln (B) + 2 ln (C^{2})]

- lo g_{10} (1.2 \cdot 1 0^{- 2})

ln (x) + ln (sin (x))

3 ln (10) - lo g_{10} (5) - lo g_{10} (2)

\frac{1}{3} ln (2 + 3)

What is 5log_{2}(k)-8log_{2}(m)+10log_{2}(n) ?
The solution to 5log_{2}(k)-8log_{2}(m)+10log_{2}(n) is log_{2}((k^5n^{10})/(m^8))