If log(12)18 = x and log(24)54 = y then prove that 5(x-y)+xy=1

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justaguide | College Teacher | (Level 2) Distinguished Educator

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It is given that log(12)18 = x and log(24)54 = y.

Let's change the base to 3

x = log 18/log 12 = (log 2 + log 9)/(log 3 + log 4)

=> (log 2 + 2)/(1 + log 4)

y = log 54/log 24 = (log 27 + log 2)/(log 3 + log 8)

=> (3 + log 2)/(1 + log 8)

5(x - y) + xy

=> 5[((log 2 + 2)/(1 + log 4)) - ((3 + log 2)/(1 + log 8))] + ((log 2 + 2)/(1 + log 4))((3 + log 2)/(1 + log 8))

=> [5(log 2 + 2)(1 + log 8) - 5(3 + log 2)(1 + log 4) + (log 2 + 2)(3 + log 2)]/(1 + log 4)(1 + log 8)

=> [5(log 2 + 2)(1 + 3log 2) - 5(3 + log 2)(1 + 2log 2) + (log 2 + 2)(3 + log 2)]/(1 + 2log 2)(1 + 3log 3)

=> [5(2 + log 2 + 3log 2*log 2 + 6log2) - 5(3 + log 2 + 6log 2 + 2*log 2*log 2) + (6 + 3log 2 + log 2*log 2 + 2log 2)]/(1 + 2log 2)(1 + 3log 3)

=> [5(2 + log 2 + 3log 2*log 2 + 6log2 - 3 - log 2 - 6log 2 - 2*log 2*log 2) + (6 + 3log 2 + log 2*log 2 + 2log 2)]/(1 + 2log 2)(1 + 3log 3)

=> [5(log 2*log 2 - 1 ) + (6 + 5log 2 + log 2*log 2)]/(1 + 2log 2)(1 + 3log 3)

=> [5log 2*log 2 - 5 + 6 + 5log 2 + log 2*log 2]/(1 + 2log 2)(1 + 3log 3)

=> [6log 2*log 2 + 1 + 5log 2]/(1 + 2log 2)(1 + 3log 3)

=> [6log 2*log 2 + 3log 2 + 2log 2 + 1]/(1 + 2log 2)(1 + 3log 3)

=> [3log 2(2log 2 + 1) + 1(2log 2 + 1)]/(1 + 2log 2)(1 + 3log 3)

=> (1+3log 2)(1 + 2log 2)/(1 + 2log 2)(1 + 3log 3)

=> 1

This proves that 5(x - y) + xy = 1

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