document.write( "Question 1205396: Show that 3+log2(9)=log2(8) \n" ); document.write( "
Algebra.Com's Answer #842168 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "\"3+%2B+log%282%2C%289%29%29+=+log%282%2C%288%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"3++=+log%282%2C%288%29%29+-+log%282%2C%289%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"3++=+log%282%2C%288%2F9%29%29\" Used log rule log(A)-log(B) = log(A/B)\r
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\n" ); document.write( "\n" ); document.write( "\"2%5E3++=+8%2F9%29\" Converting from log form to exponential form\r
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\n" ); document.write( "\n" ); document.write( "\"8++=+8%2F9%29\" We get a contradiction, so the original equation is false.\r
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\n" ); document.write( "\n" ); document.write( "Another way to show the contradiction:
\n" ); document.write( "\"3+%2B+log%282%2C%289%29%29+=+log%282%2C%288%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"3+%2B+log%282%2C%289%29%29+=+log%282%2C%282%5E3%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"3+%2B+log%282%2C%289%29%29+=+3%2Alog%282%2C%282%29%29\" Using log rule log(A^B) = B*log(A) to pull down the exponent.\r
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\n" ); document.write( "\n" ); document.write( "\"3+%2B+log%282%2C%289%29%29+=+3%2A1\" Log base b of itself is 1.\r
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\n" ); document.write( "\n" ); document.write( "\"3+%2B+log%282%2C%289%29%29+=+3\"\r
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\n" ); document.write( "\n" ); document.write( "\"log%282%2C%289%29%29+=+0\" The 3's cancel when subtracting them from both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"9+=+2%5E0\" Converting from log form to exponential form.\r
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\n" ); document.write( "\n" ); document.write( "\"9+=+1\" which is a contradiction
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