Question

If log7 – log2 + log16 – 2log3 – log(7/45) = 1 + log n find n.

Answer 1

The value of n is 4.

Explanation:

log7 – log2 + log16 – 2log3 – log(7/45) = 1 + log n

log 7 .16 /2 ( 7 ÷ 45) ^9 = 1 + log n

log 5 (8)  = 1 + log n

log 40 - log n = 1

log ( 40 / n ) = 1

log 10 (40 / n ) = 1 => 40 / n = 10

If base is 10 then,

n = 4

Hence the value of n is 4.

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Answer 2

=The value of n is 4

=>log 7 - log 2 + log 16 - 2log 3 - =>log 745 = 1 + log n

So,

=>log 7 - log 2 + log 16 - 2log 3 - (

log 7 - log 45 ) = 1 + log n

=>( log ab = log a - log b )

=>log 7 - log 2 + log 16 - log 32 -

=> log 7 + log 45 = 1 + log n

=>log ab = b log a

=>log 7 + log 16 ​ + log 45 - log 2 -

log 9 - log 7 = 1 + log n

=>log 7 × 16 ×452 × 9 × 7 = log 10

+ log n

=>log ab = log a + log b And log 10

= 1

=>log 7 × 16 ×452 × 9 × 7 = log 10n

=> log 40 = log 10 n

=>10n = 40

=> n= 4