If lo g 1 2 3 6 = x , then lo g 1 2 4 8 = a x + b , where a , b are integers. Find the value of
( b − a ) b + a
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lo g 1 2 3 6 + lo g 1 2 4 8 = lo g 1 2 ( 3 6 × 4 8 ) = lo g 1 2 1 7 2 8 = lo g 1 2 1 2 3 = 3
lo g 1 2 4 8 = 3 − lo g 1 2 3 6
lo g 1 2 4 8 = 3 − x = ( − 1 ) x + 3
on comparing we get a = − 1 , b = 3
⟹ ( b − a ) b + a = 4 2 = 1 6
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l o g 1 2 4 8 = a l o g 1 2 3 6 + b = t
1 2 t = 4 8 = ( 3 6 a ) ( 1 2 b )
2 4 3 1 = 2 2 a + 2 b 3 2 a + b
gives, b=4-1=3 and a= 2 4 − 6 =(-1).
therefore, ( a − b ) a + b = 4 2 = 1 6