document.write( "Question 114805: Solve the following:\r
\n" ); document.write( "\n" ); document.write( "log(z^2-25)-log(z+5)=log7 \n" ); document.write( "

Algebra.Com's Answer #83851 by solver91311(24713) $\"\"$ $\"About$

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$\"log%28%28z%5E2-25%29%29-log%28%28z%2B5%29%29=log7\"$ \r
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\n" ); document.write( "\n" ); document.write( "First notice that $\"z%5E2-25\"$ is the difference of two squares, so $\"log%28%28z%5E2-25%29%29=log%28%28%28z%2B5%29%28z-5%29%29%29\"$ . But we also know that the log of the product is the sum of the logs of the factors ( $\"log%28%28ab%29%29=log%28a%29%2Blog%28b%29\"$ ), so we can write:\r
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\n" ); document.write( "\n" ); document.write( " $\"log%28%28z%2B5%29%29%2Blog%28%28z-5%29%29-log%28%28z%2B5%29%29=log7\"$ \r
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\n" ); document.write( "\n" ); document.write( "Since $\"log%28%28z%2B5%29%29-log%28%28z%2B5%29%29=0\"$ , our equation reduces to:\r
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\n" ); document.write( "\n" ); document.write( " $\"log%28%28z-5%29%29=log7\"$ \r
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\n" ); document.write( "\n" ); document.write( "Since $\"log%28a%29=log%28b%29\"$ if and only if $\"a=b\"$ , we can now write:\r
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\n" ); document.write( "\n" ); document.write( " $\"z-5=7\"$
\n" ); document.write( " $\"z=12\"$ Done.\r
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