document.write( "Question 460753: im not sure of how to solve these problems,\r
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document.write( "1. log3(2x-1) + log(base 3)2 = 2
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document.write( "2. 4(^3-2x = 12
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document.write( "3. 3^x = 1/81
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document.write( "4. 9^x - 6x3^x - 27 = 0\r
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document.write( "all working would be greatly appreciated,
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document.write( "emily. \n" );
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Algebra.Com's Answer #319893 by lwsshak3(11628) ![]() You can put this solution on YOUR website! 1. log3(2x-1) + log(base 3)2 = 2 \n" ); document.write( "2. 4(^3-2x = 12 \n" ); document.write( "3. 3^x = 1/81 \n" ); document.write( "4. 9^x - 6x3^x - 27 = 0 \n" ); document.write( ".. \n" ); document.write( "1. log3(2x-1) + log(base 3)2 = 2 \n" ); document.write( "Write as single log \n" ); document.write( "log3(2x-1)(2)=2 \n" ); document.write( "convert to exponential form:(base(3) raised to log of number(2)=number((2x-1)(2) \n" ); document.write( "3^2=(2x-1)(2) \n" ); document.write( "9=4x-2 \n" ); document.write( "4x=11 \n" ); document.write( "x=11/4 \n" ); document.write( ".. \n" ); document.write( "2. 4(^3-2x = 12 \n" ); document.write( "4^3-2x=12 \n" ); document.write( "64-2x=12 \n" ); document.write( "2x=64-12=52 \n" ); document.write( "x=26 \n" ); document.write( ".. \n" ); document.write( "3. 3^x = 1/81 \n" ); document.write( "3^x=1/3^4=3^-4 \n" ); document.write( "x=-4 \n" ); document.write( ".. \n" ); document.write( "4. 9^x - 6x3^x - 27 = 0 \n" ); document.write( "I believe this equation was meant to be written as follows: \n" ); document.write( "9^x - (6)3^x - 27 = 0 \n" ); document.write( "3^2x-6*3^x-27=0 \n" ); document.write( "(3^x+3)(3^x-9)=0 \n" ); document.write( "3^x+3=0 \n" ); document.write( "3^x≠-3 (reject) \n" ); document.write( "3^x-9=0 \n" ); document.write( "3^x=9=3^2 \n" ); document.write( "x=2 \n" ); document.write( " |