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It is given that   a log 16+ b log 18 +c log 24 =   0

That is   log 16^a +  log 18^b +log 24^c   = log 1

That is log( 16^a*18^b*24^c)  =  log 1

Or 16^a*18^b*24^c  =  1. ………………….A

We have to find non zero integer values of a,b,c by trial and error that satisfy  this condition.

It can be seen that   18 and 24 or multiples of 6 .So if one number is in the numerator,the other number should be in the  denominator ,so that  some common factors between them can be cancelled out. The product of 16 or its powers and 18 or its powers would balance 24 or its powers ,so that the result is 1 

Now 16*18/24 = 12 and 16*18/ (24^2) =  1/2

Therefore { 16*18/24^2}^4 =  ( 1/2 )^4 =  1/16

Hence 16* {16*18/24^2}^4  = 16/16 =   1

That is 16^5* 18^4*24^(-8)  =1…………………..B

Comparing equation B  with equation A,we get 

a=5,b = 4 and c = -8.That is a,b  are natural numbers and c is a negative integer.

These are the minimum  values possible for a ,b,c with higher values 

being multiples of 5,4,-8 such as 10,8,-16 etc and so on.

Equation B can also be rewritten as 1/{ 16^5* 18^4*24^(-8)} =1 

Or  24^8/ { 16^5*18^4} =1  ………………….C

Comparing equation C with equation A ,we get 

c = 8, a = -5 and b= -4

Hence a^2+b^2+c^2 =   25+16+64 =  105  

It is seen from the above that when a , power of 16 is  a natural number,then  a(minimum)  =5 and  b = the power of 18  is also a natural number and b= 4

Finally if c,the power of 24 is a natural number then c =8

And a = - 5 and b= -4 and so abc = -5*-4*8 =  160 >0

 

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