document.write( "Question 559121: log8(n-3)+log8(n+4)=1 \n" ); document.write( "

Algebra.Com's Answer #363405 by lwsshak3(11628) $\"\"$ $\"About$

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log8(n-3)+log8(n+4)=1
\n" ); document.write( "log8[(n-3)(n+4)]=1
\n" ); document.write( "convert to exponential form: base(8) raised to log of number(1)=number(n-3)(n+4)
\n" ); document.write( "8^1=n^2+n-12
\n" ); document.write( "n^2+n-20=0
\n" ); document.write( "(n+5)(n-4)=0
\n" ); document.write( "n=-5 (reject, (n+4)>0
\n" ); document.write( "or
\n" ); document.write( "n=4 \n" ); document.write( "

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