If x2 + y2 =27xy,then show that log(x-y/5) =1/2 (log×+logy) ?
Answers
Answer:
ToProve:
\sf log (\frac{x-y}{5})=\frac{1}{2}[ log x + log y]log(
5
x−y
)=
2
1
[logx+logy]
\underline{\large{\sf Proof :}}
Proof:
\sf x^2+y^2=27xyx
2
+y
2
=27xy --------------(Given)
\sf x^2+y^2 = 25xy + 2xyx
2
+y
2
=25xy+2xy
\sf x^2-2xy+y^2=25xyx
2
−2xy+y
2
=25xy
{from identity,\sf(a-b)^2=a^2-2ab+b^2(a−b)
2
=a
2
−2ab+b
2
}
∴ \sf (x - y)^2 = 25xy(x−y)
2
=25xy
∴ \sf \frac{(x - y)^2}{25} = xy
25
(x−y)
2
=xy
∴ \sf (\frac{x-y}{5})^2 = xy(
5
x−y
)
2
=xy
taking square root on both the sides
\sf (\frac{x-y}{5})=(xy)^{\frac{1}{2}}(
5
x−y
)=(xy)
2
1
Now, taking log on both the sides
\sf log (\frac{x-y}{5})=log(xy)^{\frac{1}{2}}log(
5
x−y
)=log(xy)
2
1
we know, \sf log_{e}( {m}^{n} ) = n log_{e}(m)log
e
(m
n
)=nlog
e
(m)
∴ \sf log (\frac{x-y}{5})=\frac{1}{2} log(xy)log(
5
x−y
)=
2
1
log(xy)
Hence proved.