document.write( "Question 255524: log[x]=log(x+1)+1 \n" ); document.write( "
Algebra.Com's Answer #187966 by EMStelley(208)\"\" \"About 
You can put this solution on YOUR website!
\"log%28x%29+=+log%28%28x%2B1%29%29+%2B+1\"\r
\n" ); document.write( "\n" ); document.write( "The problem here is that we have two logarithms. Our first step is too subtract log(x+1) from both sides.\r
\n" ); document.write( "\n" ); document.write( "\"log%28x%29+-+log%28%28x%2B1%29%29+=+1\"\r
\n" ); document.write( "\n" ); document.write( "Now, using one of the laws of logarithms (log(a) - log(b) = log(a/b)) we have:\r
\n" ); document.write( "\n" ); document.write( "\"log%28x%2F%28x%2B1%29%29+=+1\"\r
\n" ); document.write( "\n" ); document.write( "Now we rewrite the problem in exponential form as:\r
\n" ); document.write( "\n" ); document.write( "\"10%5E1+=+x%2F%28x%2B1%29\"\r
\n" ); document.write( "\n" ); document.write( "Now multiply both sides by x+1 :\r
\n" ); document.write( "\n" ); document.write( "\"10%28x%2B1%29+=+x\"\r
\n" ); document.write( "\n" ); document.write( "\"10x%2B10+=+x\"\r
\n" ); document.write( "\n" ); document.write( "\"9x+%2B+10+=+0\"\r
\n" ); document.write( "\n" ); document.write( "\"9x+=+-10\"\r
\n" ); document.write( "\n" ); document.write( "\"x+=+-10%2F9\"\r
\n" ); document.write( "\n" ); document.write( "However, you cannot take the log of a negative number. Thus, this equation has no solution.
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