What is log_{10}(1024)-3log_{10}(2)-log_{10}(3) ?

The solution to log_{10}(1024)-3log_{10}(2)-log_{10}(3) is log_{10}(128/3)

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simplify log_{10}(1024)-3log_{10}(2)-log_{10}(3)

simplify

lo g_{10} (1024) - 3 lo g_{10} (2) - lo g_{10} (3)

lo g_{10} (\frac{128}{3})

Decimal Notation

1.63008 \dots

Solution steps

lo g_{10} (1024) - 3 lo g_{10} (2) - lo g_{10} (3)

Apply log rule:

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

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

= lo g_{10} (1024) - lo g_{10} (2^{3}) - lo g_{10} (3)

Rewrite

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

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

= lo g_{10} (1024) - (lo g_{10} (3) + lo g_{10} (2^{3}))

Apply log rule:

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

lo g_{10} (2^{3}) + lo g_{10} (3) = lo g_{10} (2^{3} \cdot 3)

= lo g_{10} (1024) - lo g_{10} (2^{3} \cdot 3)

2^{3} \cdot 3 = 24

= lo g_{10} (1024) - lo g_{10} (24)

Apply log rule:

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

lo g_{10} (1024) - lo g_{10} (24) = lo g_{10} (\frac{1024}{24})

= lo g_{10} (\frac{1024}{24})

Cancel the common factor:

8

= lo g_{10} (\frac{128}{3})

lo g_{10} (t^{3})

ln (3 x) + ln (2)

ln (\frac{n + 1}{n - 1}) - ln (n)

lo g_{2} (\frac{x}{y ^{4}})

- ln (x + 2) + ln (x - 2)

What is log_{10}(1024)-3log_{10}(2)-log_{10}(3) ?
The solution to log_{10}(1024)-3log_{10}(2)-log_{10}(3) is log_{10}(128/3)