document.write( "Question 1210448: Solve 24(2²ˣ) - 5(2ˣ⁺²) - 156 = 0, where x ∈ R
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Algebra.Com's Answer #852556 by ikleyn(53354)\"\" \"About 
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document.write( "This given equation\r\n" );
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document.write( "    \"24%2A2%5E%282x%29+\" -  \"5%2A2%5E%28x%2B2%29\"  - 156  =  0    (1)\r\n" );
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document.write( "is the same as\r\n" );
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document.write( "    \"24%5E%282x%29\"  -  \"5%2A4%2A2%5Ex\"  -  156  =  0,\r\n" );
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document.write( "or\r\n" );
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document.write( "    \"24%2A2%5E%282x%29\" - \"+20%2A2%5Ex+\" -  156  =  0.    (2)\r\n" );
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document.write( "Introduce new variable  y = \"2%5Ex\".\r\n" );
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document.write( "Then equation (2)  takes the form\r\n" );
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document.write( "    \"24y%5E2\" - \"20y\" - 156 = 0,\r\n" );
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document.write( "and we are looking for positive solutions to this equation.\r\n" );
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document.write( "Apply the quadratic formula and get the solutions to equation (2)\r\n" );
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document.write( "    y = 3  and  y = -13/6.\r\n" );
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document.write( "Only positive solution y = 3 does fit.\r\n" );
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document.write( "So, for x we have\r\n" );
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document.write( "    \"2%5Ex\" = 3,   hence,  x = \"log%282%2C%283%29%29\" = 1.584962501  (approximately).\r\n" );
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document.write( "ANSWER.  x = \"log%282%2C%283%29%29\" = 1.584962501  (approximately).\r\n" );
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\n" ); document.write( "\n" ); document.write( "This is a standard problem to solve exponential equation by introducing new variable.\r
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