Raising Logs

Algebra Level 2

$\large 5^{\log_{10} 20} 2^{\log_{10} 2} = \, ?$

2 solutions

Zee Ell
Dec 4, 2016

$5^{\log_{10}20} 2^{\log_{10}2} = 5^{1 + \log_{10}2} 2^{\log_{10}2} = 5 × 5^{\log_{10}2} 2^{\log_{10}2} =$

$= 5× (5×2)^{\log_{10}2} = 5× 10^{\log_{10}2} = 5 × 2 = \boxed {10}$

I've made some edits on the LaTeX. Writing \log will be presented as $\log$ instead of $log$ :)

Christopher Boo - 4 years, 6 months ago

Diego Emmanuel Lee Rodriguez
Dec 4, 2016

Relevant wiki: Logarithms

We know that $\log ab = \log a + \log b$ $\log_{10} 20 = \log_{10} (2*10) = \log_{10} 2 + \log_{10} 10 =\log_{10} 2 + 1$
By the definition of logarithm, we have $5^{\log_{10} \times 20} = 5^{\log_{10} 2 + 1} = 5^{log_{10} 2} \times 5$
$5^{\log_{10} 20} 2^{log_{10} 2} \\ = 5^{\log_{10} 2} \times 5 \times 2^{\log_{10}2} \\ = 10^{\log_{10} 2} \times 5 = 2 \times 5 = 10$