Question

$7log \frac{10}{9} - 2log \frac{25}{24} + 3log \frac{81}{80} = log2$

prove this.
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Answer 1

Answer:

LHS =

$\sf \: = 7 \log \frac { 16 } { 15 } + 5 \log \frac { 25 } { 24 } + 3 \log \frac { 81 } { 80 }$

$\begin{gathered}\begin{array} { l } { \sf = 7 ( \log 16 - \log 15 ) + 5 ( \log 25 - \log 24 ) + 3 ( \log 81 - \log 80 ) } \\\\ { \sf= 7 \left( \log 2 ^ { 4 } - \log ( 3 \times 5 ) \right) + 5 \left( \log 5 ^ { 2 } - \log ( 8 \times 3 ) \right) } \\ { \sf \quad + 3 \left( \log 3 ^ { 4 } - \log ( 16 \times 5 ) \right) } \end{array}\end{gathered}$

$\begin{gathered}\begin{array} { c } { \sf= 7 ( 4 \log 2 - \log 3 - \log 5 ) + 5 \left( 2 \log 5 - \log 2 ^ { 3 } - \log 3 \right) } \\ { \sf + 3(4 \log 3 - \log 2 ^ { 4 } - \log 5 ) } \\\\ { \quad = \sf \: 28 \log 2 - 7 \log 3 - 7 \log 5 + 10 \log 5 - 15 \log 2 - 5 \log 3 } \\ { \quad + \sf \: 12 \log 3 - 12 \log 2 - 3 \log 5 } \\\\ { \: \sf= 28 \log 2 - 15 \log 2 - 12 \log 2 - 7 \log 3 - 5 \log 3 + 12 \log 3 - 7 \log 5 } \\ { \quad \sf + 10 \log 5 - 3 \log 5 } \end{array}\end{gathered}$

$\begin{gathered}\begin{array} { l } {= 28 \log 2 - 27 \log 2 - 12 \log 3+ 12 \log 3 + 10 \log 5 - 10 \log 5 } \\\\ { = \log 2 } \end{array}\end{gathered}$

= RHS

$\sf \: Hence \: proved \: that \: \: \: \red{ 7 \log \frac { 16 } { 15 } + 5 \log \frac { 25 } { 24 } + 3 \log \frac { 81 } { 80 } = \log 2}$

Hence Proved.