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log base 3 ( x + 16 )= 1 + log base 3 ( 2 - X ) ----(1)
\n" ); document.write( "understanding that the base is 3 we write the above equation for convenience as
\n" ); document.write( "log( x + 16 )= 1 + log( 2 - X )
\n" ); document.write( "log( x + 16 )- log( 2 - X )=1 (change side then change sign)
\n" ); document.write( "log[(x+16)/(2-x)] = 1 (using loga-logb =log(/b) of course all for the same base)
\n" ); document.write( "That is [(x+16)/(2-x)] = 3^1
\n" ); document.write( "(using definition:log baseb(N) = p implies and is implied by N = b^p where N>0 )
\n" ); document.write( "Multiplying by (2-x) on both the sides
\n" ); document.write( "(x+16) = 3(2-x)----(2)
\n" ); document.write( "x+16 = 6 - 3x
\n" ); document.write( "x+3x = 6 -16
\n" ); document.write( "4x = -10
\n" ); document.write( "x=(-10)/4 = (-5/2)
\n" ); document.write( "Note: since x = -5/2, we must check
\n" ); document.write( "whether (x+16) and (2-x) are positive for this value of x.
\n" ); document.write( "Yes. They are positive and hence we can accept the value x =-5/2
\n" ); document.write( "Let us now verify in the equivalent form (2)
\n" ); document.write( "LHS = (x+16) =(-5/2) + 16 = 27/2
\n" ); document.write( "And RHS = 6-3x = 6+15/2 = 27/2 = LHS
\n" ); document.write( "Therefore our value is correct
\n" ); document.write( "Answer: x = (-5/2)\r
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