Log5(3÷5)+3log5(15÷2)-log5(81÷8)

Answer 1

For any base of logarithm:

log ( a ∙ b ) = log ( a ) + log ( b )

log ( a ÷ b ) = log ( a ) - log ( b )

log ( aᵐ ) = m ∙ log ( a )

In this case:

log5 (3÷5) = log5 ( 3 ) - log5 ( 5 ) = log5 ( 3 ) - 1

log5 (15 ) = log5 ( 3 ∙ 5 ) = log5 ( 3 ) + 1

log5 (15÷2 ) = log5 ( 15 ) - log5 ( 2 ) = log5 ( 3 ) + 1 - log5 ( 2 )

__________
becouse:
log5 ( 5 ) = 1
__________

log5 ( 81 ) = log5 (3⁴ ) = 4 ∙ log5 ( 3 )

log5 ( 8 ) = log5 (2³ ) = 3 ∙ log5 ( 2 )

log5 ( 81÷8 ) = log5 ( 81 ) - log5 ( 8 ) = 4 ∙ log5 ( 3 ) - 3 ∙ log5 ( 2 )

So:

log5(3÷5)+3log5(15÷2)-log5(81÷8) =

log5 ( 3 ) - 1 + 3 ∙ [ log5 ( 3 ) + 1 - log5 ( 2 ) ] - [ 4 ∙ log5 ( 3 ) - 3 ∙ log5 ( 2 ) ] =

log5 ( 3 ) - 1 + 3 log5 ( 3 ) + 3 - 3 log5 ( 2 ) - 4 log5 ( 3 ) + 3 log5 ( 2 ) =

4 ∙ 3 log5 ( 3 ) - 4 ∙ 3 log5 ( 3 ) + 2 - 3 ∙ log5 ( 2 ) + 3 ∙ log5 ( 2 ) = 2

Answer 2

assuming base 5 for all the logs, we have
(log3-log5) + 3(log15-log2) - (log81-log8)
log3-log5 + 3(log3+log5-log2) - (4log3-3log2)
log3-log5+3log3+3log5-3log2-4log3+3log2
2log5
= 2

Answer 3

The solving are too confusing it should show more steps for Better understanding.thanks it was helpful

Answer 4

Its kind of confusing...
It should be explained better

Answer 5

I to write wassce i need help

Answer 6

Intelligent