Question 884647
The general rule to follow is:
{{{ ln(a) - ln(b) = ln( a/b ) }}}
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To attempt to show why this is so, 
suppose 
{{{ a = e*e*e*e*e*e }}} ( e = 2.2718 ), or
{{{ a = e^5 }}}
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Also, suppose
{{{ b = e*e*e*e }}}, or
{{{ b = e^4 }}}
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So, now I can say:
{{{ ln(a) - ln(b) = ln(e^5) - ln(e^4) }}}
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{{{ ln(e^5) - ln(e^4) = 5 - 4 }}}
{{{ ln(e^5) - ln(e^4) = 1 }}}
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If you don't understand this last line, then
you don't really know what a log is ( natural or not ) 
and you need intensive review
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Now, to prove my definition:
{{{ ln( a/b ) = ln( e^5 / e^4 ) }}}
{{{ ln( a/b ) = ln( e^1 ) }}}
{{{ ln( a/b ) = 1 }}}
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The definition works no matter what you
plug in for {{{ a }}} or {{{ b }}}. I just made it easy.
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Now looking at your problem:
{{{ ln( x+6 ) - ln( x ) = ln(4) }}}
{{{ ln( (x+6)/x ) = ln( 4 ) }}}
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Is it clear that
{{{ ( x+6 )/x = 4 }}} ?
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The basic definition would be:
If {{{ ln(a) = ln(b) }}}, then
{{{ a = b }}}
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{{{ ( x+6 )/x = 4 }}} 
{{{ x + 6 = 4x }}}
{{{ 3x = 6 }}}
{{{ x = 2 }}}
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Check:
{{{ ln( x+6 ) - ln( x ) = ln(4) }}}
{{{ ln( 2+6 ) - ln( 2 ) = ln(4) }}}
{{{ ln( 8 ) - ln( 2 ) = ln( 4 ) }}}
{{{ ln( 8/2 ) = ln( 4 ) }}}
{{{ ln( 4 ) = ln( 4 ) }}}
OK
Hope this helps