What is the value of the following expression? $\log \left(\frac{9}{14}\right) - \log \left(\frac{15}{16}\right) + \log \left(\frac{35}{24}\right)$

### Concept & Strategy This problem tests your ability to combine multiple logarithms using product and quotient rules. Addition of logs means multiplying the arguments, while subtraction means dividing (which is equivalent to multiplying by the reciprocal). $$ \log(A) - \log(B) + \log(C) = \log\left(A \cdot \frac{1}{B} \cdot C\right) $$ ### Step-by-Step Solution **Given:** $\log \left(\frac{9}{14}\right) - \log \left(\frac{15}{16}\right) + \log \left(\frac{35}{24}\right)$ Combine the logarithmic terms into a single log expression. The subtracted term's fraction must be inverted. $\log \left( \frac{9}{14} \times \frac{16}{15} \times \frac{35}{24} \right)$ Now, simplify the large fraction by factoring it into primes to cancel common terms out between the numerator and denominator. Numerator: $9 \times 16 \times 35$ $= (3^2) \times (2^4) \times (5 \times 7)$ $= 2^4 \times 3^2 \times 5 \times 7$ Denominator: $14 \times 15 \times 24$ $= (2 \times 7) \times (3 \times 5) \times (2^3 \times 3)$ $= 2^4 \times 3^2 \times 5 \times 7$ Compare the expanded numerator and denominator. They are perfectly identical. Therefore, the fraction simplifies to $1$. Substitute back into the log expression: $\log(1)$ The logarithm of $1$ to any base is always $0$. $\log(1) = 0$ ### Exam Strategy & Shortcut Do not multiply large numbers like $9 \times 16 \times 35$. Always break them into prime factors immediately. Cancel the $2$s, $3$s, $5$s, and $7$s out visually on your paper. In most aptitude exams, expressions constructed like this are heavily rigged to cancel out to exactly $1$ or $10$. ### Common Pitfall Students often subtract fractions directly rather than applying logarithmic rules, or they forget to flip the fraction that is being subtracted. Always remember that $-\log(x/y)$ becomes $+\log(y/x)$ before you multiply the arguments. ### Final Answer **Therefore, the correct answer is 0.**