Math, asked by vishalrshetty86, 11 months ago

SHOW THAT 2log 16/15 +log 25/24 - log 32/27=0 ​WITH EXPLANATION
I HAVE A TEST TOMMOROW PLZ HELP

Answers

Answered by TanikaWaddle
9

Answer:

Step-by-step explanation:

L.H.S

2log\frac{16}{15} +log\frac{25}{24} -log\frac{32}{27}

=2log_{10}(\frac{16}{15} )+log_{10} (\frac{25}{24} )-log_{10} (\frac{32}{27})

use log rule

xlogx_{y}z=logx_{y}z^{x}

=2log_{10}(\frac{16}{15} )=log_{10}(\frac{16}{15} )^2

use log rule

logb^{(a)}+logb^{(c)} =logb^{(ac)}

[log_{10}(\frac{16}{15} )^2+log_{10} (\frac{25}{24} )]-log_{10} (\frac{32}{27})

=[log_{10}(\frac{256}{225} ) (\frac{25}{24} )]-log_{10} (\frac{32}{27})

=log_{10} (\frac{32}{27})-log_{10} (\frac{32}{27})

=0 R.H.S proved

Answered by charithpaladugu3105
3

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

Similar questions