Question 4730
{{{ln(x+3)-2=ln(x+1)}}} re-arrange to give {{{ln(x+3)-ln(x+1) = 2}}}. Now, we can use properties of logs to re-write this as:


{{{ln((x+3)/(x+1)) = 2}}}. Now we have just 1 log, we can raise everything to the power of e: this is the opposite processes to ln, just like + and - or x and divide are opposites:


{{{(x+3)/(x+1) = e^2}}} --> e is just a number like any other. Do not be scared of it.


{{{(x+3) = (e^2)*(x+1)}}}
{{{x+3 = xe^2 + e^2}}}
{{{x-xe^2 = e^2-3}}}
{{{x(1-e^2) = e^2-3}}}


so {{{x = (e^2-3)/(1-e^2)}}}


All you have done here is mess around with the order of the equation: manipulated it. Nothing too difficult. Only this you needed to know were a couple of log rules.


You can work this out on a calculator if you need to, but your answer will be an approximation, since you will round up: this algebraic version is the EXACT answer.


jon.