What is 2/3 log_{10}(2)+1/3 log_{10}(5) ?

The solution to 2/3 log_{10}(2)+1/3 log_{10}(5) is 1/3 log_{10}(20)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 2/3 log_{10}(2)+1/3 log_{10}(5)

simplify

\frac{2}{3} lo g_{10} (2) + \frac{1}{3} lo g_{10} (5)

\frac{1}{3} lo g_{10} (20)

Decimal Notation

0.43367 \dots

Solution steps

\frac{2}{3} lo g_{10} (2) + \frac{1}{3} lo g_{10} (5)

Apply log rule:

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

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

= lo g_{10} (2^{\frac{2}{3}}) + \frac{1}{3} lo g_{10} (5)

Apply log rule:

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

\frac{1}{3} lo g_{10} (5) = lo g_{10} (5^{\frac{1}{3}})

= lo g_{10} (2^{\frac{2}{3}}) + lo g_{10} (5^{\frac{1}{3}})

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{10} (2^{\frac{2}{3}}) + lo g_{10} (5^{\frac{1}{3}}) = lo g_{10} (2^{\frac{2}{3}} \cdot 5^{\frac{1}{3}})

= lo g_{10} (2^{\frac{2}{3}} \cdot 5^{\frac{1}{3}})

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

Rewrite as

= lo g_{10} ((2^{2} \cdot 5)^{\frac{1}{3}})

Apply log rule

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

assuming

x \geq 0

= \frac{1}{3} lo g_{10} (2^{2} \cdot 5)

2^{2} = 4

= \frac{1}{3} lo g_{10} (4 \cdot 5)

Multiply the numbers:

4 \cdot 5 = 20

= \frac{1}{3} lo g_{10} (20)

0.5 ln (\frac{100}{2}) + 0.5 ln (\frac{100}{2})

10 lo g_{10} (\frac{i}{i _{0}})

\frac{ln ( 15 )}{2 ln ( 2 )} - 7

\frac{1}{10} ln (\frac{11}{15})

ln (7 x + 4) + ln (x + 8)

What is 2/3 log_{10}(2)+1/3 log_{10}(5) ?
The solution to 2/3 log_{10}(2)+1/3 log_{10}(5) is 1/3 log_{10}(20)