What is 1/2 log_{10}(24)-log_{10}(4) ?

The solution to 1/2 log_{10}(24)-log_{10}(4) is log_{10}((sqrt(3))/(sqrt(2)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 1/2 log_{10}(24)-log_{10}(4)

simplify

\frac{1}{2} lo g_{10} (24) - lo g_{10} (4)

lo g_{10} (\frac{3}{2})

Decimal Notation

0.08804 \dots

Solution steps

\frac{1}{2} lo g_{10} (24) - lo g_{10} (4)

lo g_{10} (4) = 2 lo g_{10} (2)

= \frac{1}{2} lo g_{10} (24) - 2 lo g_{10} (2)

\frac{1}{2} lo g_{10} (24) = lo g_{10} (2 4^{\frac{1}{2}})

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

= lo g_{10} (2 4^{\frac{1}{2}}) - lo g_{10} (2^{2})

Apply log rule:

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

lo g_{10} (2 4^{\frac{1}{2}}) - lo g_{10} (2^{2}) = lo g_{10} (\frac{2 4 ^{\frac{1}{2}}}{2 ^{2}})

= lo g_{10} (\frac{2 4 ^{\frac{1}{2}}}{2 ^{2}})

Cancel

\frac{2 4 ^{\frac{1}{2}}}{2 ^{2}} : \frac{3 ^{\frac{1}{2}}}{2 ^{\frac{1}{2}}}

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

Apply exponent rule:

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

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

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

Apply exponent rule:

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

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

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

lo g_{b} (x^{3} \frac{y ^{2}}{z})

ln (\frac{4 x}{7 y ^{2}})

lo g_{5} (\frac{25 b ^{5} y ^{2} - 6}{y ( x + 6 ) ^{2}})

lo g_{2} (\frac{p ^{5}}{q ^{9}})

What is 1/2 log_{10}(24)-log_{10}(4) ?
The solution to 1/2 log_{10}(24)-log_{10}(4) is log_{10}((sqrt(3))/(sqrt(2)))