Question 115701
{{{log(5,3x+15)-log(5,3x)=1}}}


You need to use three of the rules for logarithms for this problem.


1) {{{log(b,x^a)=a*log(b,x)}}}, and
2) {{{log(b,f(x))+log(b,g(x))=log(b,f(x)*g(x))}}}
3) {{{log(b,x)=y}}} is equivalent to {{{b^y=x}}}


Using the first rule, you can write:


{{{log(5,3x+15)+log(5,3x^(-1))=1}}}, notice that the minus sign in the original equation became an exponent of -1.


{{{log(5,3x+15)+log(5,(1/3x))=1}}} (using the rule {{{x^(-1)=1/x}}})


Now applying rule 2) we can write:
{{{log(5,(3x+15)*(1/3x))=1}}}


Now, using rule 3) we can write:
{{{5^1=(3x+15)*(1/3x)}}}


Simplify and solve for x
{{{15x=3x+15}}}
{{{12x=15}}}
{{{4x=5}}}
{{{x=5/4}}}.  Done.