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

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

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

simplify

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

lo g_{10} (\frac{5}{4})

Decimal Notation

0.09691 \dots

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

lo g_{10} (5) - lo g_{10} (4^{3}) = lo g_{10} (\frac{5}{4 ^{3}})

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

4^{3} = 64

= 4 lo g_{10} (2) + lo g_{10} (\frac{5}{64})

Apply log rule:

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

4 lo g_{10} (2) = lo g_{10} (2^{4})

= lo g_{10} (2^{4}) + lo g_{10} (\frac{5}{64})

Apply log rule:

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

lo g_{10} (2^{4}) + lo g_{10} (\frac{5}{64}) = lo g_{10} (2^{4} \cdot \frac{5}{64})

= lo g_{10} (2^{4} \cdot \frac{5}{64})

2^{4} \cdot \frac{5}{64} = \frac{5}{4}

= lo g_{10} (\frac{5}{4})

ln (3 x^{2} - 9 x) + ln (\frac{1}{3 x})

- lo g_{10} (3.5 \cdot 1 0^{11})

ln (k e^{- bk} b^{2})

\frac{1}{2} ln b^{2} + 1 - \frac{1}{2} ln a^{2} + 1

lo g_{1.8} (\frac{5.07}{3.92}) + 51.7 - 51.6

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