if a=3,b=5,c=7 then prove the above equation
Answers
Answer:
\large{\green{\underline{\underline{\mathcal{Given}}}}}
Given
\bf{a=3}a=3
\bf{b=5}b=5
\bf{c=7}c=7
\begin{gathered}\small{\orange{ \tt{\boxed{\underline{L.H.S= \frac{5 log_{10} \sqrt{b} + 5 log_{10}3 - log_{10}( {a}^{2} {c}^{2} \sqrt{ac} ) }{ log_{10}( \frac{ab}{c} )} \div \frac{5}{2}}}}}}\\\sf{= \frac{5 log_{10} {5}^{ \frac{1}{2} } + 5 log_{10}3 - log_{10}( {3}^{2} {7}^{2} \sqrt{3.7} ) }{ log_{10}( \frac{3.5}{7} )} \div \frac{5}{2}}\\\\\sf{= \frac{\frac{5}{2} log_{10}5 + 5 log_{10}3 - log_{10} {(3.7)}^{ \frac{5}{2} } }{ log_{10} \frac{15}{7} } \div \frac{5}{2}}\\\\\sf{= \frac{\frac{5}{2} log_{10}5 + 5 log_{10} \sqrt{9} - \frac{5}{2} log_{10}21}{ log_{10}\frac{15}{7} } \div \frac{5}{2}}\\\\\\\sf{ = \frac{\frac{5}{2} log_{10}5 + 5 log_{10} {9}^{ \frac{1}{2} } - \frac{5}{2} log_{10}21}{ log_{10}\frac{15}{7} } \div \frac{5}{2}}\\\\\\\sf{= \frac{\frac{5}{2} log_{10}5 + \frac{5}{2} log_{10} 9 - \frac{5}{2} log_{10}21}{ log_{10}\frac{15}{7} } \div \frac{5}{2}}\\\\\\\sf{ = \frac{ \frac{5}{2} log_{10} \frac{5 \times 9}{21} }{ log_{10} \frac{15}{7} } \div \frac{5}{2}}\\\\\\\sf{ = \frac{ \frac{5}{2} ( log_{10}\frac{15}{7} ) }{( log_{10} \frac{15}{7}) } \div \frac{5}{2}} \\\\\\\sf{= \frac{5}{2} \times \frac{2}{5}} \\\\\\\sf{= 1}\\\\\\\tt{\orange{\boxed{\underline{ = R.H.S}}}}\end{gathered}
L.H.S=
log
10
(
c
ab
)
5log
10
b
+5log
10
3−log
10
(a
2
c
2
ac
)
÷
2
5
=
log
10
(
7
3.5
)
5log
10
5
2
1
+5log
10
3−log
10
(3
2
7
2
3.7
)
÷
2
5
=
log
10
7
15
2
5
log
10
5+5log
10
3−log
10
(3.7)
2
5
÷
2
5
=
log
10
7
15
2
5
log
10
5+5log
10
9
−
2
5
log
10
21
÷
2
5
=
log
10
7
15
2
5
log
10
5+5log
10
9
2
1
−
2
5
log
10
21
÷
2
5
=
log
10
7
15
2
5
log
10
5+
2
5
log
10
9−
2
5
log
10
21
÷
2
5
=
log
10
7
15
2
5
log
10
21
5×9
÷
2
5
=
(log
10
7
15
)
2
5
(log
10
7
15
)
÷
2
5
=
2
5
×
5
2
=1
=R.H.S
\begin{gathered}\:\:\:\:\small\underline{\green{\boxed{\tt{\: \frac{5 log_{10} \sqrt{b} + 5 log_{10}3 - log_{10}( {a}^{2} {c}^{2} \sqrt{ac} ) }{ log_{10}( \frac{ab}{c} )} \div \frac{5}{2}\:=\:\bf{1}}}}}\end{gathered}
log
10
(
c
ab
)
5log
10
b
+5log
10
3−log
10
(a
2
c
2
ac
)
÷
2
5
=1