document.write( "Question 167780: Can you please help me solve this? This is what I have so far.
\n" ); document.write( "log5(x+3)+log5(x-3)=3
\n" ); document.write( "log5{(x+3)(x-3)}=3
\n" ); document.write( "(x+3)(x-3)=3^5
\n" ); document.write( "x^2-9=243\r
\n" ); document.write( "\n" ); document.write( "How do I finish this?\r
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Algebra.Com's Answer #123680 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
You're doing great. However, the third step should be \"%28x%2B3%29%28x-3%29=5%5E3\" (note the switch on the right side) which makes the fourth step \"x%5E2-9=125\"\r
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\n" ); document.write( "\n" ); document.write( "Note: remember, if \"log%28b%2C%28y%29%29=x\", then \"b%5Ex=y\". So the base of the log should be the base of the exponential expression\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-9=125\" Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2=125%2B9\" Add 9 to both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2=134\" Combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"x=sqrt%28134%29\" Take the square root of both sides. Note: only the positive square root is considered since you CANNOT take the log of a negative number.\r
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\n" ); document.write( "\n" ); document.write( "So the solution is \"x=sqrt%28134%29\"
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