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

The solution to 4log_{10}(2)-2log_{10}(3)-log_{10}(4) is log_{10}(4/9)

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

simplify

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

lo g_{10} (\frac{4}{9})

Decimal Notation

- 0.35218 \dots

Solution steps

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

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

Apply log rule:

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

2 lo g_{10} (3) = lo g_{10} (3^{2})

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

Apply log rule:

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

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

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

\frac{2 ^{4}}{3 ^{2}} = \frac{16}{9}

= lo g_{10} (\frac{16}{9}) - lo g_{10} (4)

Apply log rule:

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

lo g_{10} (\frac{16}{9}) - lo g_{10} (4) = lo g_{10} (\frac{\frac{16}{9}}{4})

= lo g_{10} (\frac{\frac{16}{9}}{4})

\frac{\frac{16}{9}}{4} = \frac{4}{9}

= lo g_{10} (\frac{4}{9})

ln (8) - ln (6)

ln (8) + ln (x)

lo g_{2} (a) + lo g_{2} (25) - lo g_{4} ((5 a)^{2})

lo g_{a} (mn)

ln (2 x + 1) + ln (x + 2)

What is 4log_{10}(2)-2log_{10}(3)-log_{10}(4) ?
The solution to 4log_{10}(2)-2log_{10}(3)-log_{10}(4) is log_{10}(4/9)