7 l o g ( 16 15 ) + 5 l o g ( 25 24 ) + 3 l o g ( 81 80 ) is equal to - Options: A . &nbsp 0 B . &nbsp 1 C . &nbsp log 2 D . &nbsp log 3

Answer: Option C : C 7 l o g ( 16 15 ) + 5 l o g ( 25 24 ) + 3 l o g ( 81 80 ) ∵ a l o g x = l o g x a ⇒ l o g ( 16 15 ) 7 + l o g ( 25 24 ) 5 + l o g ( 81 80 ) 3 ∵ l o g x + l o g y = l o g ( x × y ) ⇒ l o g ( 16 7 15 7 . 25 5 24 5 . 81 3 80 3 ) = l o g 2

7 log (1615)+5 log(2524)+3 log(8180) is equal to -

Discussion Forum : Inequalities Modulus And Logarithms

Question -

$7 l o g (\frac{16}{15}) + 5 l o g (\frac{25}{24}) + 3 l o g (\frac{81}{80})$ is equal to -

Options:

A . 0

B . 1

C . log 2

D . log 3

Answer: Option C
:
C

7 l o g (\frac{16}{15}) + 5 l o g (\frac{25}{24}) + 3 l o g (\frac{81}{80})

∵ a l o g x = l o g x^{a}

\Rightarrow l o g (\frac{16}{15})^{7} + l o g (\frac{25}{24})^{5} + l o g (\frac{81}{80})^{3}

∵ l o g x + l o g y = l o g (x \times y)

\Rightarrow l o g (\frac{16^{7}}{15^{7}} . \frac{25^{5}}{24^{5}} . \frac{81^{3}}{80^{3}})

= l o g 2

Was this answer helpful ?

Next Question

Discussion Forum : Inequalities Modulus And Logarithms

Submit Your Solution hear: