document.write( "Question 829852: find x
\n" ); document.write( "1) log (x-1) to base 2 + log (x+1) to base 2 = 3
\n" ); document.write( "2) log (4-2x)=log(x+1)
\n" ); document.write( "3) log (9-2x) to base 1/3=log1/3(x-6) \n" ); document.write( "
Algebra.Com's Answer #500289 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
find x
\n" ); document.write( "1) \"log%282%2C%28x-1%29%29+%2B+log%282%2C%28x%2B1%29%29\" = 3
\n" ); document.write( "we can write this as single log
\n" ); document.write( "\"log%282%2C%28%28x-1%29%28x%2B1%29%29%29\" = 3
\n" ); document.write( "\"log%282%2C%28x%5E2-1%29%29\" = 3
\n" ); document.write( "write the exponent equiv of logs
\n" ); document.write( "\"%28x%5E2+-+1%29+=+2%5E3\"
\n" ); document.write( "\"%28x%5E2+-+1%29+=+8\"
\n" ); document.write( "\"x%5E2+-+1+-+8+=+0\"
\n" ); document.write( "\"x%5E2+-+9\" = 0
\n" ); document.write( "Factors as the difference of squares
\n" ); document.write( "(x+3)(x-3) = 0
\n" ); document.write( "x = +3, the only valid solution, can't have the log of a neg value
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "2) log (4-2x)=log(x+1)
\n" ); document.write( "if the logs are equal the expressions are equal, just write it:
\n" ); document.write( "4 - 2x = x + 1
\n" ); document.write( "4 - 1 = x + 2x
\n" ); document.write( "3 = 3x
\n" ); document.write( "x = 1
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "3) \"log%281%2F3%2C%289-2x%29%29\" =\"log%281%2F3%2C%28x-6%29%29\"
\n" ); document.write( "If the expressions are equal of logs with same base
\n" ); document.write( "9 - 2x = x - 6
\n" ); document.write( "9 + 6 = x + 2x
\n" ); document.write( "15 = 3x
\n" ); document.write( "x = 5
\n" ); document.write( "However if you substitute in the original problem, we end up with -1 on each
\n" ); document.write( "side, no log of neg number is allowed, so no solution \n" ); document.write( "
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