If $\log_{3} x + \log_{9} x^2 + \log_{27} x^3 = 9$, then $x$ equals

### Concept & Formula This problem is solved efficiently by normalizing the bases. We use the powerful property that allows us to factor out exponents from both the base and the argument of a logarithm. $$ \log_{a^n} (x^m) = \frac{m}{n} \log_a x $$ ### Step-by-Step Solution * **Given:** An equation with three logarithms having different bases ($3$, $9$, $27$) but related arguments ($x$, $x^2$, $x^3$). $$ \log_3 x + \log_9 x^2 + \log_{27} x^3 = 9 $$ * **Calculation:** Express the bases 9 and 27 as powers of 3 to unify the equation. $9 = 3^2$ $27 = 3^3$ $$ \log_3 x + \log_{3^2} x^2 + \log_{3^3} x^3 = 9 $$ * Apply the property $\log_{a^n} (x^m) = \frac{m}{n} \log_a x$ to simplify the second and third terms. For the second term: $\log_{3^2} x^2 = \frac{2}{2} \log_3 x = 1 \log_3 x$. For the third term: $\log_{3^3} x^3 = \frac{3}{3} \log_3 x = 1 \log_3 x$. * Substitute these simplified terms back into the original equation. $$ \log_3 x + \log_3 x + \log_3 x = 9 $$ * Add the identical terms together. $$ 3 \log_3 x = 9 $$ * Divide both sides by 3. $$ \log_3 x = 3 $$ * Convert the logarithmic equation to an exponential equation to solve for $x$. $$ x = 3^3 $$ $$ x = 27 $$ ### Exam Strategy & Shortcut Look for symmetry. The terms are $\log_{3^1} x^1$, $\log_{3^2} x^2$, and $\log_{3^3} x^3$. The power of the base perfectly matches the power of the argument in every single term. This means the powers instantly cancel out, leaving you with three identical $\log_3 x$ terms. So, $3 \log_3 x = 9$, giving $\log_3 x = 3$, leading straight to $3^3 = 27$. ### Common Pitfall A lengthy and error-prone approach is using the standard change-of-base formula $\left(\frac{\log x^2}{\log 9}\right)$ and converting everything to base 10. While valid, it introduces complex fractions and algebraic manipulation that consumes valuable exam time. Master the exponent factorization property for bases and arguments instead. ### Final Answer **Therefore, the correct answer is 27.**