document.write( "Question 172047: Hi, \r
\n" ); document.write( "\n" ); document.write( "I am having a problem solving: \r
\n" ); document.write( "\n" ); document.write( "log(z^2-25)-log(z+5)=log7 \r
\n" ); document.write( "\n" ); document.write( "The farthest I got was to re-write the problem to: log(z^2-25)(z+5)=7
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Algebra.Com's Answer #127081 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
Close, but no cigar this time.\r
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\n" ); document.write( "\n" ); document.write( "\"log%28%28z%5E2-25%29%29-log%28%28z%2B5%29%29=log%28%287%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "Remember the rules:\r
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\n" ); document.write( "\n" ); document.write( "The sum of the logs is the log of the PRODUCT.\r
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\n" ); document.write( "\n" ); document.write( "The difference of the logs is the log of the QUOTIENT.\r
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\n" ); document.write( "\n" ); document.write( "So you should have written:\r
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\n" ); document.write( "\n" ); document.write( "\"log%28%28%28z%5E2-25%29%2F%28z%2B5%29%29%29=log%28%287%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "From here you should be able to recognize that \"z%5E2-25=%28z%2B5%29%28z-5%29\" so \"log%28%28%28z%5E2-25%29%2F%28z%2B5%29%29%29=+log%28%28z-5%29%29+=+log%28%287%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "Furthermore, \"log%28b%2C%28a%29%29=log%28b%2C%28c%29%29\" if and only if \"a=c\", so:\r
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\n" ); document.write( "\n" ); document.write( "\"z-5=7\"\r
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\n" ); document.write( "\n" ); document.write( "\"z=12\"
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