Prove
Asked by ac5744100
26th July 2018,
11:10 AM
Answered by Expert
Answer:
alpha = log (base 12) 18 = log 18/log 12
beta = log (base 24) 54 = log 54/log 24
So (alpha)(beta) + 5(alpha - beta)
= log 18 log 54/(log 12 log 24) + 5(log 18/log 12 - log 54/log 24)
= (log 18 log 54 + 5 log 18 log 24 - 5 log 54 log 12)/(log 12 log 24)
log 18 log 54
= (log 2 + 2 log 3)(log 2 + 3 log 3)
= (log 2)^2 + 5 log 2 log 3 + 6(log 3)^2
5 log 18 log 24
= 5(log 2 + 2 log 3)(3 log 2 + log 3)
= 15(log 2)^2 + 35 log 2 log 3 + 10 (log 3)^2
5 log 54 log 12
= 5(log 2 + 3 log 3)(2 log 2 + log 3)
= 10 (log 2)^2 + 35 log 2 log 3 + 15(log 3)^2
So
log 18 log 54 + 5 log 18 log 24 - 5 log 54 log 12
= (1 + 15 - 10) (log 2)^2 + (5 + 35 - 35) log 2 log 3 + (6 + 10 - 15) (log 3)^2
= 6 (log 2)^2 + 5 log 2 log 3 + (log 3)^2
and log 12 log 24
= (2 log 2 + log 3)(3 log 2 + log 3)
= 6(log 2)^2 + 5 log 2 log 3 + (log 3)^2
These two expressions were the numerator and denominator respectively.
It follows that
(log 18 log 54 + 5 log 18 log 24 - 5 log 54 log 12)/(log 12 log 24) = 1
beta = log (base 24) 54 = log 54/log 24
So (alpha)(beta) + 5(alpha - beta)
= log 18 log 54/(log 12 log 24) + 5(log 18/log 12 - log 54/log 24)
= (log 18 log 54 + 5 log 18 log 24 - 5 log 54 log 12)/(log 12 log 24)
log 18 log 54
= (log 2 + 2 log 3)(log 2 + 3 log 3)
= (log 2)^2 + 5 log 2 log 3 + 6(log 3)^2
5 log 18 log 24
= 5(log 2 + 2 log 3)(3 log 2 + log 3)
= 15(log 2)^2 + 35 log 2 log 3 + 10 (log 3)^2
5 log 54 log 12
= 5(log 2 + 3 log 3)(2 log 2 + log 3)
= 10 (log 2)^2 + 35 log 2 log 3 + 15(log 3)^2
So
log 18 log 54 + 5 log 18 log 24 - 5 log 54 log 12
= (1 + 15 - 10) (log 2)^2 + (5 + 35 - 35) log 2 log 3 + (6 + 10 - 15) (log 3)^2
= 6 (log 2)^2 + 5 log 2 log 3 + (log 3)^2
and log 12 log 24
= (2 log 2 + log 3)(3 log 2 + log 3)
= 6(log 2)^2 + 5 log 2 log 3 + (log 3)^2
These two expressions were the numerator and denominator respectively.
It follows that
(log 18 log 54 + 5 log 18 log 24 - 5 log 54 log 12)/(log 12 log 24) = 1
Answered by Expert
26th July 2018,
12:58 PM
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