SOLUTION: if x=log to the base 2a,a ,y=log to the base 3a,2a ,z=log to the base 4a,3a ,prove that x*y*z+1=2*y*z
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Question 199819: if x=log to the base 2a,a ,y=log to the base 3a,2a ,z=log to the base 4a,3a ,prove that x*y*z+1=2*y*z
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
Given,,,x= log(2a)a,,,,y -log(3a) 2a,,,,z+ log(4a)3a
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to prove x*y*z+1 = 2*y*z
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Remember log(b)x= logx/logb
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loga/log2a * log2a/log3a * log 3a /log4a ) +1 = 2* log 2a/log3a * log 3a/log4a
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reduce by eliminating like num and den
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(loga / log4a) +1 = 2* log2a/log4a
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Remember 2 loga = loga^2,,,,and a/a =1, loga/loga =1
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(loga/log4a) + log4a/log4a = log (2a)^2 / log 4a
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Remember adding fractions with common den ,,,,and log(a*b) = loga + log b
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(loga +log4a ) / log4a = log 4a^2 / log 4a
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log (a*4a) / log 4a = log 4a^2 /log4a
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log 4a^2 / log 4a = log 4a^2 / log 4a
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To check, let a = 2,, check original and answer, all 1 1/3 = 1 1/3 ,,,ok
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