Question

(Log 49+log 343-log 2401)÷(log16807-log 7)?

Answer 1

`1/4`

Explanation:

As,
`49 = 7^2`
`343 = 7^3`
`2401 = 7^4`
`16807 = 7^5`

Rewriting,

`(log49 + log343 - log2401)/(log16807 - log7)`

`=> (log7^2 + log7^3 - log7^4)/(log7^5 - log7)`

Using logarithmic properties,
`=> (2log7 + 3log7 - 4log7)/(5log7 - log7)`

`=> log7/(4log7)`

Hence,
the value the comes out to be `1/4`

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HOPE THIS HELPS !!!!