1. Logarithms

If $a = \log_{10} \frac{10}{9} , b = \log_{10} \frac{25}{24} and
c = \log_{10} \frac{81}{80}$

prove that $\log_{10} 2 = 7a-2b+3c$

2. Originally Posted by ecogreen

If $a = \log_{10} \frac{10}{9} , b = \log_{10} \frac{25}{24} and
c = \log_{10} \frac{81}{80}$

prove that $\log_{10} 2 = 7a-2b+3c$

$7a-2b+3c=7\log\frac{10}{9}-2\log\frac{25}{24}+3\log\frac{81}{80}$ $=7(\log 2+\log 5-2\log 3)-2(2\log 5-3\log 2-\log 3)+3(4\log 3-4\log 2 -\log 5)$ , and now open up parenthese and you're done.

Logarithms' properties used: $\log\frac{A}{B}=\log A-\log B\,,\,\log(AB)=\log A+\log B\,,\,\log(A^n)=n\log A$

Tonio