Question

Prove that :-
log2+16log16/15+12log25/24+7log81/80 =1​

Answer 1

Answer:

Change everything to log 2, log 3, and log 5.

log 2 = log 2

16 log (16/15)

= 16 (log 16 - log 15)

= 16 (log (2^4) - log (3 x 5))

= 16 ((4 log 2 - (log 3 + log 5))

= 64 log 2 - 16 log 3 - 16 log 5

12 log (25/24)

= 12 (log 25 - log 24)

= 12 (log (5^2) - log ((2^3) (3)))

= 12 (2 log 5 - (log (2^3) + log 3))

= 12 (2 log 5 - (3 log 2 + log 3))

= - 36 log 2 - 12 log 3 + 24 log 5

7 log (81/80)

= 7 (log 81 - log 80)

= 7 ( log 3^4 - log ((2^4) (5))

= 7 (4 log 3 - (log 2^4 + log 5))

= 7 (3 log 3 - 4 log 2 - log 5)

= - 28 log 2 + 28 log 3 - 7 log 5

Add them = log 2 + log 5 = log 10 = 1

Thank you.

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Answer 2

Step-by-step explanation:

We can write

n log a = log aⁿ and

log m + log n + log p = log (mnp)

Hence

log 2+ 16 log (16/15) + 12 log (25/24) + 7 log (81/80)

= log 2 + log (16/15)^16 + log (25/24)^12 + log (81/80)^7

= log 2 + log {2⁴/(3*5)}^16 + log{5²/(3*2³)}^12 + log{3⁴/(2⁴*5)}^7

= log [ 2* {2^64/(3^16*5^16)}*{5^24/(3^12*2^36)}*{3^28/(2^28*5^7) }]

= log ( 2*5)

= log 10

= 1.