If $a = b^2 = c^3 = d^4$, then the value of $\log_a (abcd)$ would be

### Concept & Strategy The strategy is to express all variables in terms of a single variable (in this case, $a$, since it is the base of the logarithm) and then use the logarithmic product rule to expand the expression. $$ \log_x(p \cdot q \cdot r \cdot s) = \log_x p + \log_x q + \log_x r + \log_x s $$ ### Step-by-Step Solution * **Given:** The equality chain $a = b^2 = c^3 = d^4$. We need to find $\log_a (abcd)$. * **Calculation:** First, express $b$, $c$, and $d$ in terms of $a$ using fractional exponents. $b^2 = a \Rightarrow b = a^{\frac{1}{2}}$ $c^3 = a \Rightarrow c = a^{\frac{1}{3}}$ $d^4 = a \Rightarrow d = a^{\frac{1}{4}}$ * Expand the target logarithmic expression using the product rule. $$ \log_a(abcd) = \log_a a + \log_a b + \log_a c + \log_a d $$ * Substitute the expressions for $b$, $c$, and $d$ that we found in terms of $a$. $$ \log_a a + \log_a(a^{\frac{1}{2}}) + \log_a(a^{\frac{1}{3}}) + \log_a(a^{\frac{1}{4}}) $$ * Use the power rule ($\log_x(y^n) = n \log_x y$) to bring the fractional exponents to the front. Also, note that $\log_a a = 1$. $$ 1 + \frac{1}{2}\log_a a + \frac{1}{3}\log_a a + \frac{1}{4}\log_a a $$ * Since $\log_a a = 1$ in every term, it simplifies cleanly to fractions. $$ 1 + \frac{1}{2}(1) + \frac{1}{3}(1) + \frac{1}{4}(1) $$ $$ 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} $$ ### Exam Strategy & Shortcut An even faster method skips expanding the logarithm first. Multiply the variables directly: $abcd = a \cdot a^{\frac{1}{2}} \cdot a^{\frac{1}{3}} \cdot a^{\frac{1}{4}}$. When multiplying terms with the same base, add their exponents: $abcd = a^{(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})}$. Taking $\log_a$ of this expression instantly drops the base $a$, leaving only the exponent: $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$. ### Common Pitfall Some students incorrectly look at $a = b^2 = c^3 = d^4$ and try to substitute coefficients inside the log, such as assuming $\log_a b = 2$ instead of $\frac{1}{2}$. Always be careful when isolating the variable; $b^2 = a$ means taking the square root, hence the fractional power. ### Final Answer **Therefore, the correct answer is $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$.**