document.write( "Question 1194878: Please help me solve this question:log2{√(x³-2x²+x)}=1+log2(x-1) \n" ); document.write( "
Algebra.Com's Answer #827179 by MathTherapy(10552)\"\" \"About 
You can put this solution on YOUR website!
Please help me solve this question:log2{√(x³-2x²+x)}=1+log2(x-1)
\n" ); document.write( "
\r\n" );
document.write( " ------ Multiplying by LCD, 2 \r\n" );
document.write( " ------ Subtracting \"2+%2A+log+%282%2C+%28x++-++1%29%29\" from both sides\r\n" );
document.write( "\r\n" );
document.write( "                    \"matrix%281%2C3%2C+%28x%28x++-++1%29%5E2%29%2F%28x++-++1%29%5E2%2C+%22=%22%2C+2%5E2%29\" ------ Converting to EXPONENTIAL form \r\n" );
document.write( "At this point, with x - 1 in the denominator, the equation's \"x\" values can be ANY REAL NUMBER, EXCEPT 1.\r\n" );
document.write( "                      \"matrix%281%2C3%2C+x%28x+-+1%29%5E2%2C+%22=%22%2C+4%28x+-+1%29%5E2%29\" ------- Cross-multiplying\r\n" );
document.write( "            \r\n" );
document.write( "Therefore, x = 4 and also, x = 1 (DOUBLE-ROOT)\r\n" );
document.write( "However, since x CANNOT = 1 (\"x+%3C%3E+1\", as stated above), the ONLY solution is: x = 4. \r\n" );
document.write( "\r\n" );
document.write( "Just to make sure that no errors were made, and since this involves LOGARITHMS and square roots, you'll need to CHECK to make sure that x = 4 is \r\n" );
document.write( "NOT an EXTRANEOUS solution.
\n" ); document.write( "
\n" );