Question 4711
log(x+2)-logx=(x+5) 


Use the second law of logarithms {{{log M - log N = log (M / N) }}}to express the left side as a quotient:

{{{log ((x+2)) - log x = log ((x+5)) }}}
{{{log ((x+2)/ (x)) = log ((x+ 5)) }}}


log M = log N means that M=N.

Therefore {{{ (x+2) / x = (x+5) }}}


Multiply both sides of the equation by x:
{{{x+ 2 = x(x+ 5) }}}
{{{x+2 = x^2 + 5x }}}


Set equal to zero:
{{{ 0 = x^2 + 5x - x - 2}}}
{{{ 0 = x^2  + 4x - 2}}}


Are you sure you copied this right?  They "usually" factor at this point.  Nevertheless,

Solve by quadratic formula or completing the square.  Completing the square is actually easier.

{{{x^2  + 4x - 2 = 0}}}  Add +2 to each side and place a blank space on each side of the equation:

{{{ x^2 + 4x + _____ = 2 + _____}}}


Take half of the x coefficient, and square it.  Half of 4 is 2, and 2^2 = 4.  Add +4 to each side of the equation on the blank lines above:

{{{ x^2 + 4x + 4 = 2 + 4}}}
{{{ (x+ 2) ^2 = 6}}}


Take square root of each side of the equation:
{{{x+2 = 0 +- sqrt (6) }}}


Subract 2 from each side:
{{{x= -2 +- sqrt (6)}}}


Of course, you must reject the value {{{-2 - sqrt (6)}}}, since it makes the log of a negative.  The final answer is  {{{-2 + sqrt (6) }}}


Are you sure you didn't copy this wrong??


R^2 at SCC