What is 12+1/2 log_{10}(7)-log_{10}(2) ?

The solution to 12+1/2 log_{10}(7)-log_{10}(2) is 12+log_{10}((7^{1/2})/2)

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simplify 12+1/2 log_{10}(7)-log_{10}(2)

simplify

12 + \frac{1}{2} lo g_{10} (7) - lo g_{10} (2)

12 + lo g_{10} (\frac{7 ^{\frac{1}{2}}}{2})

Decimal Notation

12.12151 \dots

Solution steps

12 + \frac{1}{2} lo g_{10} (7) - lo g_{10} (2)

Apply log rule:

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

\frac{1}{2} lo g_{10} (7) = lo g_{10} (7^{\frac{1}{2}})

= 12 + lo g_{10} (7^{\frac{1}{2}}) - lo g_{10} (2)

Apply log rule:

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

lo g_{10} (7^{\frac{1}{2}}) - lo g_{10} (2) = lo g_{10} (\frac{7 ^{\frac{1}{2}}}{2})

= 12 + lo g_{10} (\frac{7 ^{\frac{1}{2}}}{2})

lo g_{2} (x) + 4 lo g_{2} (y)

2 (450 ln (30) - 200 - 50 ln (10))

ln (- 5 a e^{- 5 a})

ln (e \cdot x)

\frac{1}{3} ln ∣ x ∣ + \frac{2}{3} ln ∣ x + 3 ∣ + C

What is 12+1/2 log_{10}(7)-log_{10}(2) ?
The solution to 12+1/2 log_{10}(7)-log_{10}(2) is 12+log_{10}((7^{1/2})/2)