Question

(ii) Prove that
(log 8 + log 27 + log 125)divided by
1 + log 3
= 3.​

Answer 1

Step-by-step explanation:

$log(8) = log_{2}(3) \\ \\ log(27) = log_{3}(3) \\ \\ log(125) = log_{5}(3) \\ \\ log_{3}(3) = 1 \\ \\ \frac{1 + log_{2}(3) + log_{5}(3) }{1 + log(3) } \\ \\ \frac{1 + log_{2}(3) + log_{3}(5) }{ log_{3}(3) + log_{3}(1) } \\ \\ log_{a}(x) + log_{a}(y) = log_{a}(xy) \\ \\ \frac{1 + log_{2}(3) + log_{5}(3) }{ log_{3}(3) = 1 } = 1 + log_{2}(3) + log_{5}(3)$

$log_{a}(a)=1$

$log_{2}(3)+log_{5}(3)=3$

so 1+3=4

see the attachment it is done by calculator and it is roughly equal to 4

3.75= approx(4)

swipe the answer left and right to see the remaining portion

Answer 2

Solution :

We have to prove that (log 8 + log 27 + log 125) divided by (1+log 3) is equal to 3.

The base is 10 for all cases.

log 8 = log(2)³ = 3 log 2

log 27 = log(3)³ = 3 log 3

log 125 = log(5)³ = 3 log 5

log 8 + log 27 + log 125

>> 3 log 2 + 3 log 3 + 3 log 5

>> 3( log 2 + log 3 + log 5)

>> 3[ (log 2 + log 5) + log 3]

>> 3[ log10 + log 3]

log 10 is 1 ( as base is 10)

>> 3( 1 + log 3) .

When this is divided by (1+log3) , we get 3 .

Hence Proved

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