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81^x=9^(x^2-3)
\n" ); document.write( "[(9)^2]^x = 9^(x^2-3)----(1)\r
\n" ); document.write( "\n" ); document.write( "(9)^(2x) = 9^(x^2-3)
\n" ); document.write( "This implies
\n" ); document.write( "(2x) = (x^2-3) (bases same implies powers equal)
\n" ); document.write( "0 = x^2-2x-3
\n" ); document.write( "That is x^2-2x-3 = 0
\n" ); document.write( "(x-3)(x+1) = 0
\n" ); document.write( "(x-3)= 0 gives x=3
\n" ); document.write( "(x+1) = 0 gives x = -1
\n" ); document.write( "Answer: x = 3 and x = -1
\n" ); document.write( "Verification: x=3 in (1)
\n" ); document.write( "LHS = [(9)^2]^x = [(9)^2]^3 = [(9)^6]
\n" ); document.write( "RHS = 9^(x^2-3)= 9^(3^2-3)= [(9)^6]=LHS
\n" ); document.write( "x=-1 in (1)
\n" ); document.write( "LHS = [(9)^2]^(-1) = [(9)^(-2)]
\n" ); document.write( "RHS = 9^[(-1)^2-3)= 9^(1-3)= [(9)^(-2)]=LHS
\n" ); document.write( "Therefore our answers are correct\r
\n" ); document.write( "\n" ); document.write( "Note: Given x^2-3x+x-3 = 0 how to factorise the LHS expression.
\n" ); document.write( "[splitting the middle term into two parts so that their sum is the middle term and their product is the product of the square term and the constant term. Here (-2x) = (-3x)+(x) and (-3x)X(x) = -3x^2 = (x^2)X(-3)]
\n" ); document.write( "(x^2-3x)+(x-3) = 0
\n" ); document.write( "x(x-3)+1(x-3)=0
\n" ); document.write( "xp+p= 0 where p = (x-3)
\n" ); document.write( "p(x+1) = 0
\n" ); document.write( "(x-3)(x+1) = 0
\n" ); document.write( "(x-3)= 0 gives x=3
\n" ); document.write( "(x+1) = 0 gives x = -1
\n" ); document.write( "Answer: x = 3 and x = -1 \r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "