document.write( "Question 1203918: 2log5x+3log3y=8
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Algebra.Com's Answer #839836 by ikleyn(52914)\"\" \"About 
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document.write( "Introduce new variables  u = log(5x),  v = log(3y).\r\n" );
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document.write( "Then the original systems of equations takes the form\r\n" );
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document.write( "    2u + 3v = 8    (1)\r\n" );
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document.write( "    6u + 2v = 2    (2)\r\n" );
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document.write( "Solve using elimination\r\n" );
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document.write( "    6u + 9v = 24\r\n" );
document.write( "    6u + 2v =  2\r\n" );
document.write( "---------------------------Subtract\r\n" );
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document.write( "         7v = 22  --->  v = \"22%2F7\".\r\n" );
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document.write( "Then from equation (1) we have\r\n" );
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document.write( "    2u + 66/7 = 8\r\n" );
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document.write( "    2u = 8 - 66/7 = -10/7 --->  u = \"-5%2F7\".\r\n" );
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document.write( "Now, log(5x) = u = \"-5%2F7\"  --->  5x = 10^(-5/7)  --->  x = (1/5)*10^(-5/7).\r\n" );
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document.write( "     log(3y) = v = 22/7  --->  3y = 10^(22/7)  --->  y = (1/3)*10^(22/7).\r\n" );
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document.write( "ANSWER.  x = (1/5)*10^(-5/7),  y = (1/3)*10^(22/7).\r\n" );
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\n" ); document.write( "\n" ); document.write( "The method to solve such problem is to introduce new variables
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