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

The solution to log_{10}(12)+1/2 log_{10}(7)-log_{10}(2) is log_{10}(6sqrt(7))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

simplify

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

lo g_{10} (67)

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

lo g_{a} (x) + lo g_{a} (y) = lo g_{a} (x y)

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

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

Apply log rule:

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

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

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

Divide the numbers:

\frac{12}{2} = 6

= lo g_{10} (6 \cdot 7^{\frac{1}{2}})

Apply exponent rule:

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

7^{\frac{1}{2}} = 7

= lo g_{10} (67)

lo g_{10} (3) + lo g_{10} (4)

lo g_{10} (\frac{6 ( x + 4 ) ^{3}}{x ^{5}})

ln (5 e^{6})

- \frac{1}{4} lo g_{2} (\frac{1}{4}) - \frac{3}{4} lo g_{2} (\frac{3}{4})

ln (5 e^{x})

What is log_{10}(12)+1/2 log_{10}(7)-log_{10}(2) ?
The solution to log_{10}(12)+1/2 log_{10}(7)-log_{10}(2) is log_{10}(6sqrt(7))