Question 33989
i assume that the logs are all base10? If so,


2log(x+2)+log4 = log x + 4 log3
2log(x+2) = log x + 4 log3 - log4
2log(x+2) - log x = 4 log3 - log4
{{{ log(x+2)^2 - log x = log3^4 - log4 }}}
{{{ log((x+2)^2/x) = log(3^4/4) }}}


Now, raise both terms to power 10, to cancel the logs:
{{{ (x+2)^2/x = 3^4/4 }}}
{{{ (x+2)^2/x = 81/4 }}}
{{{ 4(x+2)^2 = 81x }}}
{{{ 4(x^2+4x+4) = 81x }}}
{{{ 4x^2+16x+16 = 81x }}}
{{{ 4x^2-65x+16 = 0 }}}


This is just a simple quadratic now, to either factorise if you can spot it or use the quadratic formula if not.


{{{ x = (-b +- sqrt(b^2-4ac))/(2a) }}}
{{{ x = (-(-65) +- sqrt((-65)^2-4(4)(16)))/(2(4)) }}}


Working this through gives x=16 or x=0.25


jon.