prove that 7 log16 /15+5 log25/24+3log81/80=log2
Answers
Answer:
Hence proved that 7 \log \frac { 16 } { 15 } + 5 \log \frac { 25 } { 24 } + 3 \log \frac { 81 } { 80 } = \log 27log
15
16
+5log
24
25
+3log
80
81
=log2
Solution:
Given,
On simplifying the LHS, we get
LHS =
= 7 \log \frac { 16 } { 15 } + 5 \log \frac { 25 } { 24 } + 3 \log \frac { 81 } { 80 }=7log
15
16
+5log
24
25
+3log
80
81
\begin{gathered}\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}\end{gathered}
=7(log16−log15)+5(log25−log24)+3(log81−log80)
=7(log2
4
−log(3×5))+5(log5
2
−log(8×3))
+3(log3
4
−log(16×5))
\begin{gathered}\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}\end{gathered}
=7(4log2−log3−log5)+5(2log5−log2
3
−log3)
+3(4log3−log2
4
−log5)
=28log2−7log3−7log5+10log5−15log2−5log3
+12log3−12log2−3log5
=28log2−15log2−12log2−7log3−5log3+12log3−7log5
+10log5−3log5
\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}
=28log2−27log2−12log3+12log3+10log5−10log5
=log2
= RHS
Hence proved that 7 \log \frac { 16 } { 15 } + 5 \log \frac { 25 } { 24 } + 3 \log \frac { 81 } { 80 } = \log 27log
15
16
+5log
24
25
+3log
80
81
=log2