SHOW THAT 2log 16/15 +log 25/24 - log 32/27=0 WITH EXPLANATION
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Answers
Answer:
Step-by-step explanation:
L.H.S
use log rule
use log rule
Answer:
0
Step-by-step explanation:
L.H.S
2log\frac{16}{15} +log\frac{25}{24} -log\frac{32}{27}2log
15
16
+log
24
25
−log
27
32
=2log_{10}(\frac{16}{15} )+log_{10} (\frac{25}{24} )-log_{10} (\frac{32}{27})=2log
10
(
15
16
)+log
10
(
24
25
)−log
10
(
27
32
)
use log rule
xlogx_{y}z=logx_{y}z^{x}xlogx
y
z=logx
y
z
x
=2log_{10}(\frac{16}{15} )=log_{10}(\frac{16}{15} )^2=2log
10
(
15
16
)=log
10
(
15
16
)
2
use log rule
logb^{(a)}+logb^{(c)} =logb^{(ac)}logb
(a)
+logb
(c)
=logb
(ac)
[log_{10}(\frac{16}{15} )^2+log_{10} (\frac{25}{24} )]-log_{10} (\frac{32}{27})[log
10
(
15
16
)
2
+log
10
(
24
25
)]−log
10
(
27
32
)
=[log_{10}(\frac{256}{225} ) (\frac{25}{24} )]-log_{10} (\frac{32}{27})=[log
10
(
225
256
)(
24
25
)]−log
10
(
27
32
)
=log_{10} (\frac{32}{27})-log_{10} (\frac{32}{27})=log
10
(
27
32
)−log
10
(
27
32
)
=0 R.H.S proved=0R.H.Sproved