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How to solve:
log_(3) (x+6) + log_(3)(x-6) = 4

User OliverD
by
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1 Answer

3 votes

Collapse the logarithms into one:


\log_3(x+6)+\log_3(x-6)=\log_3((x+6)(x-6))=4

Then write both sides as powers of 3:


3^(\log_3((x+6)(x-6)))=3^4


(x+6)(x-6)=81

Simplify the left side and solve for
x:


x^2-36=81


x^2=117


x=\pm√(117)=\pm3√(13)

But we're not done yet. Notice that both
x+6 and
x-6 are negative when
x=-3√(13). We can't take the logarithm of a negative number (the result isn't real-valued, anyway), so we throw this solution out, and we're left with just


x=3√(13)

User Bagljas
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