What is 3log_{10}(4)-5log_{10}(2)+log_{10}(6) ?

The solution to 3log_{10}(4)-5log_{10}(2)+log_{10}(6) is log_{10}(12)

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simplify 3log_{10}(4)-5log_{10}(2)+log_{10}(6)

simplify

3 lo g_{10} (4) - 5 lo g_{10} (2) + lo g_{10} (6)

lo g_{10} (12)

Decimal Notation

1.07918 \dots

Solution steps

3 lo g_{10} (4) - 5 lo g_{10} (2) + lo g_{10} (6)

Apply log rule:

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

5 lo g_{10} (2) = lo g_{10} (2^{5})

= 3 lo g_{10} (4) - lo g_{10} (2^{5}) + lo g_{10} (6)

Apply log rule:

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

lo g_{10} (6) - lo g_{10} (2^{5}) = lo g_{10} (\frac{6}{2 ^{5}})

= 3 lo g_{10} (4) + lo g_{10} (\frac{6}{2 ^{5}})

\frac{6}{2 ^{5}} = \frac{3}{16}

= 3 lo g_{10} (4) + lo g_{10} (\frac{3}{16})

Apply log rule:

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

3 lo g_{10} (4) = lo g_{10} (4^{3})

= lo g_{10} (4^{3}) + lo g_{10} (\frac{3}{16})

Apply log rule:

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

lo g_{10} (4^{3}) + lo g_{10} (\frac{3}{16}) = lo g_{10} (4^{3} \cdot \frac{3}{16})

= lo g_{10} (4^{3} \cdot \frac{3}{16})

4^{3} \cdot \frac{3}{16} = 12

= lo g_{10} (12)

ln (x e^{x^{2}})

ln (x) + ln (8)

\frac{1}{4} ln (3)

ln (\frac{3}{10})

lo g_{10} (\frac{x ^{9}}{y ^{4}})

What is 3log_{10}(4)-5log_{10}(2)+log_{10}(6) ?
The solution to 3log_{10}(4)-5log_{10}(2)+log_{10}(6) is log_{10}(12)