Question

prove that 7log 16/15+5log 25/24+3log 81/80=log2

Answer 1

Answer:

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

Solution:

Given,  

On simplifying the LHS, we get  

LHS =

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

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

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

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

= RHS

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

Answer 2

Answer:

hence proved that  7log 16/15+5log 25/24+3log 81/80=log2

Step-by-step explanation:

7log 16/15+5log 25/24+3log 81/80=log2

   = log[(16/15)^7] + log[(25/24)^5] + log[(81/80)^3]

    = log{[(16^7)*(25^5)*(81^3)] / [(15^7)*(24^5)*(80^3)]}

   = log{[(2^28)*(5^10)*(3^12)] / [(3^7)*(5^7)*(2^15)*(3^5)*(2^12)*(5^3]}

   = log{[(2^28)*(3^12)*(5^10)] / [(2^27)*(3^12)*(5^10)}

    = log(2)

hope it helps you..!!!