Question

express the logarithms of the following as the sum of the logarithm 1) 35×46, 2) 235×437, 3) 2437×3568​

Answer 1

Answer:

1)

Since

We have to find log of 35×46

concept used

log(a×b) = log(a) + log(b)

log(35×46) = log(35) + log(46)

                  = log(5×7) + log(2×23)

                  = log(5) + log(7) + log(2) + log(23)

So

log(35×46) = log(5) + log(7) + log(2) + log(23)

2)

log(235×437) = log(235) + log(437) =  log(5×47) + log(437)

                      = log(5) + log(47) + log(437)

So

log(235×437) = log(5) + log(47) + log(437)                

3)

log(2437×3568) = log(2437) + log(3568)

since

3568 = 2×2×2×2×223

So

log(2437) + log(3568) = log(2437) + log(2×2×2×2×223)

                                     = log(2437) + log(2)+log(2)+log(2)+log(2)+log(223)

So

log(2437×3568) = log(2437) + log(2) + log(2) + log(2) + log(2) + log(223)

Answer 2

Answer:  The required sums are

$(1)~\log 35+\log 46,~~\(2)~\log 235+\log 437,~~(3)~\log 2437+\log 3568.$

Step-by-step explanation:  We are given to express the logarithms of the following as the sum of the logarithms :

1) 35×46,

2) 235×437,

3) 2437×3568​.

We know that

for any two positive integers a and b, we have the following logarithmic property :

$\log(a+b)=\log a+\log b.$

So. we get

$(1)~\log(35\times46)=\log 35+\log 46,\\\\\\(2)~\log(235\times 437)=\log 235+\log 437,\\\\\\(3)~\log(2437\times 3568)=\log 2437+\log 3568.$

Thus, the required sums are

$(1)~\log 35+\log 46,~~\(2)~\log 235+\log 437,~~(3)~\log 2437+\log 3568.$