What is 1/2 log_{10}(3)-1/2 log_{10}(12) ?

The solution to 1/2 log_{10}(3)-1/2 log_{10}(12) is -log_{10}(2)

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simplify 1/2 log_{10}(3)-1/2 log_{10}(12)

simplify

\frac{1}{2} lo g_{10} (3) - \frac{1}{2} lo g_{10} (12)

- lo g_{10} (2)

Decimal Notation

- 0.30102 \dots

Solution steps

\frac{1}{2} lo g_{10} (3) - \frac{1}{2} lo g_{10} (12)

\frac{1}{2} lo g_{10} (3) = lo g_{10} (3^{\frac{1}{2}})

- \frac{1}{2} lo g_{10} (12) = - lo g_{10} (1 2^{\frac{1}{2}})

= lo g_{10} (3^{\frac{1}{2}}) - lo g_{10} (1 2^{\frac{1}{2}})

Apply log rule:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

lo g_{10} (3^{\frac{1}{2}}) - lo g_{10} (1 2^{\frac{1}{2}}) = lo g_{10} (\frac{3 ^{\frac{1}{2}}}{1 2 ^{\frac{1}{2}}})

= lo g_{10} (\frac{3 ^{\frac{1}{2}}}{1 2 ^{\frac{1}{2}}})

Cancel

\frac{3 ^{\frac{1}{2}}}{1 2 ^{\frac{1}{2}}} : \frac{1}{2}

= lo g_{10} (\frac{1}{2})

\frac{1}{2} = 2^{- 1}

= lo g_{10} (2^{- 1})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), x > 0

lo g_{10} (2^{- 1}) = - 1 \cdot lo g_{10} (2)

= - 1 \cdot lo g_{10} (2)

Apply rule:

1 \cdot a = a

= - lo g_{10} (2)

ln (\frac{x}{e})

- \frac{1}{2} (ln (3) - ln (2)) + \frac{5}{6}

lo g_{10} (\frac{m ^{2} + n ^{2}}{100000})

ln (x + 5) + ln (x - 5)

What is 1/2 log_{10}(3)-1/2 log_{10}(12) ?
The solution to 1/2 log_{10}(3)-1/2 log_{10}(12) is -log_{10}(2)