Question 147145
{{{-3log(10,(x+2))+log(10,(5))=0}}} Start with the given equation



{{{-3log(10,(x+2))=-log(10,(5))}}} Subtract {{{log(10,(5))}}} from both sides.



{{{log(10,(x+2)^(-3))=-log(10,(5))}}} Rewrite the left side using the identity  {{{y*log(b,(x))=log(b,(x^y))}}}



{{{log(10,(x+2)^(-3))=log(10,(5)^(-1))}}} Rewrite the right side using the identity  {{{y*log(b,(x))=log(b,(x^y))}}}



{{{(x+2)^(-3)=(5)^(-1)}}} Since the logs are equal with the same base, this means that arguments are equal



{{{1/(x+2)^(3)=1/(5)^(1)}}} Flip the fractions



{{{5^1=(x+2)^(3)}}} Cross multiply.



{{{5=(x+2)^(3)}}} Simplify.



{{{root(3,5)=x+2}}} Take the cube root of both sides.



{{{root(3,5)-2=x}}} Subtract 2 from both sides.



{{{-2+root(3,5)=x}}} Rearrange the terms.




So the answer is {{{x=-2+root(3,5)}}}