What is 2log_{m}(5)-log_{m}(2)+log_{m}(4) ?

The solution to 2log_{m}(5)-log_{m}(2)+log_{m}(4) is log_{m}(50)

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simplify 2log_{m}(5)-log_{m}(2)+log_{m}(4)

simplify

2 lo g_{m} (5) - lo g_{m} (2) + lo g_{m} (4)

lo g_{m} (50)

Solution steps

2 lo g_{m} (5) - lo g_{m} (2) + lo g_{m} (4)

Apply log rule:

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

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

= lo g_{m} (5^{2}) - lo g_{m} (2) + lo g_{m} (4)

Apply log rule:

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

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

= lo g_{m} (\frac{5 ^{2}}{2}) + lo g_{m} (4)

5^{2} = 25

= lo g_{m} (\frac{25}{2}) + lo g_{m} (4)

Apply log rule:

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

lo g_{m} (\frac{25}{2}) + lo g_{m} (4) = lo g_{m} (4 \cdot \frac{25}{2})

= lo g_{m} (4 \cdot \frac{25}{2})

\frac{25}{2} \cdot 4 = 50

= lo g_{m} (50)

3 (ln (e^{3 x} (x + 1)) - ln (2 x^{2}))

\frac{1}{2} ln (\frac{17}{3})

- lo g_{10} (7 \cdot 1 0^{- 6})

ln (\frac{10}{6})

lo g_{5} (a) + \frac{1}{3} lo g_{5} (b)

What is 2log_{m}(5)-log_{m}(2)+log_{m}(4) ?
The solution to 2log_{m}(5)-log_{m}(2)+log_{m}(4) is log_{m}(50)