Question 133332
{{{log(10,(x))+log(10,(x+3))=1}}} Start with the given equation



{{{log(10,(x(x+3)))=1}}} Combine the logs using the identity {{{log(b,(A))+log(b,(B))=log(b,(A*B))}}}



{{{log(10,(x^2+3x))=1}}} Distribute



{{{10^(log(10,(x^2+3x)))=10^1}}} Raise both sides as exponents with bases of 10. This eliminates the logs



{{{x^2+3x=10}}} Simplify



{{{x^2+3x-10=0}}}  Subtract 10 from both sides. 




{{{(x+5)(x-2)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{x+5=0}}} or  {{{x-2=0}}} 


{{{x=-5}}} or  {{{x=2}}}    Now solve for x in each case



So our possible answers are 


 {{{x=-5}}} or  {{{x=2}}} 




However, since you cannot take the log of a negative number, the only solution is {{{x=2}}}



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Answer:

So our answer is {{{x=2}}}