document.write( "Question 948119: How do I solve algebraically for all values of x:\r
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document.write( "log(x+3)*(2x+3)+log(x+3)*(x+5)=2\r
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document.write( "In this case, (x+3) is the base. \n" );
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Algebra.Com's Answer #578646 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! How do I solve algebraically for all values of x: \n" ); document.write( "log(x+3)*(2x+3)+log(x+3)*(x+5)=2 \n" ); document.write( "In this case, (x+3) is the base. \n" ); document.write( "*** \n" ); document.write( " \n" ); document.write( "convert to exponential form: base(x+3) raised to log of number(2)=numberz((2x+3)(x+5)) \n" ); document.write( "(x+3)^2=(2x+3)(x+5) \n" ); document.write( "x^2+6x+9=2x^2+13x+15 \n" ); document.write( "x^2+7x+6=0 \n" ); document.write( "(x+6)(x+1)=0 \n" ); document.write( "x=-6 (reject, (x+3)<0 \n" ); document.write( "or \n" ); document.write( "x=1 \n" ); document.write( " |