What is log_{10}(583)-1/2 log_{10}(150) ?

The solution to log_{10}(583)-1/2 log_{10}(150) is log_{10}((583)/(5sqrt(6)))

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simplify log_{10}(583)-1/2 log_{10}(150)

simplify

lo g_{10} (583) - \frac{1}{2} lo g_{10} (150)

lo g_{10} (\frac{583}{5 6})

Decimal Notation

1.67762 \dots

Solution steps

lo g_{10} (583) - \frac{1}{2} lo g_{10} (150)

Apply log rule:

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

- \frac{1}{2} lo g_{10} (150) = - lo g_{10} (15 0^{\frac{1}{2}})

= lo g_{10} (583) - lo g_{10} (15 0^{\frac{1}{2}})

Apply log rule:

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

lo g_{10} (583) - lo g_{10} (15 0^{\frac{1}{2}}) = lo g_{10} (\frac{583}{15 0 ^{\frac{1}{2}}})

= lo g_{10} (\frac{583}{15 0 ^{\frac{1}{2}}})

15 0^{\frac{1}{2}} = 5 \cdot 6^{\frac{1}{2}}

= lo g_{10} (\frac{583}{5 \cdot 6 ^{\frac{1}{2}}})

Apply exponent rule:

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

6^{\frac{1}{2}} = 6

= lo g_{10} (\frac{583}{5 6})

ln (3.89 \cdot 1 0^{- 34})

2 lo g_{2} (3) + lo g_{2} (x)

ln (\frac{12}{\frac{7.02}{0.0445}})

What is log_{10}(583)-1/2 log_{10}(150) ?
The solution to log_{10}(583)-1/2 log_{10}(150) is log_{10}((583)/(5sqrt(6)))