Question 114805
{{{log((z^2-25))-log((z+5))=log7}}}


First notice that {{{z^2-25}}} is the difference of two squares, so {{{log((z^2-25))=log(((z+5)(z-5)))}}}.  But we also know that the log of the product is the sum of the logs of the factors ({{{log((ab))=log(a)+log(b)}}}), so we can write:


{{{log((z+5))+log((z-5))-log((z+5))=log7}}}


Since {{{log((z+5))-log((z+5))=0}}}, our equation reduces to:


{{{log((z-5))=log7}}}


Since {{{log(a)=log(b)}}} if and only if {{{a=b}}}, we can now write:


{{{z-5=7}}}
{{{z=12}}}   Done.