Question 177448
1.  {{{ln(sqrt(x -8))=5}}}


First note that {{{sqrt(a) = a^(1/2)}}}, so:  {{{ln((x -8)^(1/2))=5}}}


Now use the rule that {{{log(b,(a^n)) = n*log(b,(a))}}} for any {{{n}}}, any {{{a>0}}}, and any {{{b>0}}}, so:  {{{(1/2)ln(x-8) = 5}}} → {{{ln(x -8) = 10}}}


Now use the definition {{{log(b,(x)) = y}}} if and only if {{{b^y=x}}}


So:  {{{e^10 = x - 8}}} → {{{x = e^10 + 8}}}, which is the exact answer.  Use your calculator to get a numerical approximation if that is what you need.


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{{{ ln(x) + ln(x + 3) = 1 }}}


First use the rule:  The sum of the logs equals the log of the product, {{{log(b,(a[1])) + log(b,(a[2])) = log(b,(a[1]a[2]))}}}, so: {{{ln(x) + ln(x + 3) = ln(x(x+3)) = ln(x^2 + 3x) = 1}}}.


Now use the definition {{{log(b,(x)) = y}}} if and only if {{{b^y=x}}}, so:


{{{ln(x^2 + 3x) = 1}}} → {{{e^1 = e = x^2 + 3x}}} → {{{x^2 + 3x - e = 0}}}


Use the quadratic formula:  {{{x = (-3 +- sqrt( 3^2-4(1)(-e) ))/2(1) }}} 


{{{x = (-3 +- sqrt( 9+4e ))/2 }}}  Again, if you need a numerical approximation, use your calculator.


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Use the rule:  {{{log(b,(a)) = log(b, (c))}}} if and only if {{{a = c}}}


So: {{{x - 6 = 2x + 1}}}.  You should be able to handle the rest of this one on your own.