Numerical: logarithms MCQ - Practice Questions with Answers

Write 9 as 3^2 and 27 as 3^3. Then log_9 27 means log base 3^2 of 3^3, which equals 3/2. Similarly, log_27 9 means log base 3^3 of 3^2, which equals 2/3. Adding them gives 3/2 + 2/3 = 9/6 + 4/6 = 13/6. The values 5/6, 1/2, and 9/5 result from reversing, multiplying, or simplifying the logarithmic terms incorrectly.

Use \(1/\log_a b=\log_b a\). The expression becomes \(\log_{72}3+\log_{72}4+\log_{72}6\), which combines as \(\log_{72}(3\times4\times6)\). Since \(3\times4\times6=72\), the value is \(\log_{72}72=1\). Therefore 1 is correct. The value cannot be 0 because the product inside the combined logarithm is not 1. It cannot be 2 or 5 because the combined logarithm has the same base and argument, giving exactly one, not a higher power.

Since log_10 100 equals 2, the expression inside the outer logarithm becomes 5 × 2 = 10. Now log_100 10 is 1/2, because 100^(1/2) = 10. Cubing this value gives (1/2)^3 = 1/8. The value 1 ignores the fractional logarithm, 1/16 would arise from using the wrong power, and 1/10 confuses the argument 10 with the final cubed value.