$\log_{10} \frac{26}{51} + \log_{10} \frac{119}{91} - \log_{10} \frac{13}{32} - \log_{10} \frac{64}{39}$ is equal to

### Concept & Strategy This problem tests the application of the product and quotient rules of logarithms to combine a string of logarithmic terms into a single expression. Addition of logs translates to multiplication of their arguments, while subtraction translates to division (or multiplication by the reciprocal). $$ \log(A) + \log(B) - \log(C) = \log\left(\frac{A \cdot B}{C}\right) $$ ### Step-by-Step Solution **Given:** $\log_{10} \left(\frac{26}{51}\right) + \log_{10} \left(\frac{119}{91}\right) - \log_{10} \left(\frac{13}{32}\right) - \log_{10} \left(\frac{64}{39}\right)$ Using the log properties, we can combine all terms into a single base-$10$ logarithm. Terms being added are multiplied in the numerator; terms being subtracted are divided (which means we multiply by their reciprocals). Combine the arguments: $\log_{10} \left( \frac{26}{51} \times \frac{119}{91} \times \frac{32}{13} \times \frac{39}{64} \right)$ Now, break down all numbers into prime factors to easily cancel terms out. Numerator terms: $26 = 2 \times 13$ $119 = 7 \times 17$ $32 = 2^5$ $39 = 3 \times 13$ Combined Numerator = $2 \times 13 \times 7 \times 17 \times 2^5 \times 3 \times 13 = 2^6 \times 3 \times 7 \times 13^2 \times 17$ Denominator terms: $51 = 3 \times 17$ $91 = 7 \times 13$ $13 = 13$ $64 = 2^6$ Combined Denominator = $3 \times 17 \times 7 \times 13 \times 13 \times 2^6 = 2^6 \times 3 \times 7 \times 13^2 \times 17$ Notice that the combined numerator is exactly identical to the combined denominator. Therefore, the entire fraction simplifies to $1$. Substitute back into the log: $\log_{10} (1)$ Since the logarithm of $1$ to any base is always $0$: $\log_{10} (1) = 0$ ### Exam Strategy & Shortcut When faced with a long string of fractional log additions and subtractions, write one giant fraction immediately. Flip the fractions that have a minus sign in front of their log and multiply everything. Cancel common factors aggressively before doing any actual multiplication. In competitive exams, these fractions are specifically designed to perfectly cancel out to $1$, $10$, or $100$. ### Common Pitfall The most frequent error is mishandling the subtracted logs. Students sometimes subtract the fractions directly or forget to invert the fractions before multiplying. Remember: $-\log(a/b)$ is equivalent to $+\log(b/a)$. Always flip the subtracted terms before combining. ### Final Answer **Therefore, the correct answer is 0.**