Logarithms: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 2)

Here we need log base 2 of (4)^3. The power rule log_a(x^p) = p x log_a(x) applies here. First find log base 2 of 4. Since 4 = 2^2, log base 2 of 4 = 2. Now apply the power rule: 3 x 2 = 6. So log base 2 of (4)^3 = 6. Check: 2^6 = (2^2)^3 = 4^3 = 64.

Here we need log base 3 of (81)^4. The power rule log_a(x^p) = p x log_a(x) applies here. First find log base 3 of 81. Since 81 = 3^4, log base 3 of 81 = 4. Now apply the power rule: 4 x 4 = 16. So log base 3 of (81)^4 = 16. Check: 3^16 = (3^4)^4 = 81^4 = 43046721.

Here we need (log base 7 of 16807) + (log base 7 of 16807). The law log_a(x) + log_a(y) = log_a(xy) applies here. So the sum equals log base 7 of (16807 x 16807). Now 16807 x 16807 = 282475249. Since 16807 = 7^5 and 16807 = 7^5, we get 282475249 = 7^(5+5) = 7^10. By definition, log base 7 of 7^10 = 10. So the answer is 10.

Here we need (log base 5 of 15625) + (log base 5 of 25). The law log_a(x) + log_a(y) = log_a(xy) applies here. So the sum equals log base 5 of (15625 x 25). Now 15625 x 25 = 390625. Since 15625 = 5^6 and 25 = 5^2, we get 390625 = 5^(6+2) = 5^8. By definition, log base 5 of 5^8 = 8. So the answer is 8.

Here we need (log base 5 of 3125) + (log base 5 of 15625). The law log_a(x) + log_a(y) = log_a(xy) applies here. So the sum equals log base 5 of (3125 x 15625). Now 3125 x 15625 = 48828125. Since 3125 = 5^5 and 15625 = 5^6, we get 48828125 = 5^(5+6) = 5^11. By definition, log base 5 of 5^11 = 11. So the answer is 11.

Here we need (log base 4 of 4096) + (log base 4 of 4096). The law log_a(x) + log_a(y) = log_a(xy) applies here. So the sum equals log base 4 of (4096 x 4096). Now 4096 x 4096 = 16777216. Since 4096 = 4^6 and 4096 = 4^6, we get 16777216 = 4^(6+6) = 4^12. By definition, log base 4 of 4^12 = 12. So the answer is 12.

Here we need (log base 5 of 25) + (log base 5 of 25). The law log_a(x) + log_a(y) = log_a(xy) applies here. So the sum equals log base 5 of (25 x 25). Now 25 x 25 = 625. Since 25 = 5^2 and 25 = 5^2, we get 625 = 5^(2+2) = 5^4. By definition, log base 5 of 5^4 = 4. So the answer is 4.

Here we need (log base 4 of 64) + (log base 4 of 16). The law log_a(x) + log_a(y) = log_a(xy) applies here. So the sum equals log base 4 of (64 x 16). Now 64 x 16 = 1024. Since 64 = 4^3 and 16 = 4^2, we get 1024 = 4^(3+2) = 4^5. By definition, log base 4 of 4^5 = 5. So the answer is 5.

Here we need (log base 8 of 64) + (log base 8 of 512). The law log_a(x) + log_a(y) = log_a(xy) applies here. So the sum equals log base 8 of (64 x 512). Now 64 x 512 = 32768. Since 64 = 8^2 and 512 = 8^3, we get 32768 = 8^(2+3) = 8^5. By definition, log base 8 of 8^5 = 5. So the answer is 5.

Here we need (log base 5 of 15625) + (log base 5 of 15625). The law log_a(x) + log_a(y) = log_a(xy) applies here. So the sum equals log base 5 of (15625 x 15625). Now 15625 x 15625 = 244140625. Since 15625 = 5^6 and 15625 = 5^6, we get 244140625 = 5^(6+6) = 5^12. By definition, log base 5 of 5^12 = 12. So the answer is 12.

Here log base x of 161051 = 5. Converting to exponential form: log_x(a) = k means x^5 = a. So x^5 = 161051. To solve for x, take the 5th root of both sides. So x = 161051^(1/5) = 11. Check: 11^5 = 161051.

Here log base x of 256 = 4. Converting to exponential form: log_x(a) = k means x^4 = a. So x^4 = 256. To solve for x, take the 4th root of both sides. So x = 256^(1/4) = 4. Check: 4^4 = 256.

Here log base x of 16807 = 5. Converting to exponential form: log_x(a) = k means x^5 = a. So x^5 = 16807. To solve for x, take the 5th root of both sides. So x = 16807^(1/5) = 7. Check: 7^5 = 16807.

Here log base x of 100 = 2. Converting to exponential form: log_x(a) = k means x^2 = a. So x^2 = 100. To solve for x, take the square root of both sides. So x = 100^(1/2) = 10. Check: 10^2 = 100.

Here log base x of 81 = 2. Converting to exponential form: log_x(a) = k means x^2 = a. So x^2 = 81. To solve for x, take the square root of both sides. So x = 81^(1/2) = 9. Check: 9^2 = 81.

Here log base x of 64 = 2. Converting to exponential form: log_x(a) = k means x^2 = a. So x^2 = 64. To solve for x, take the square root of both sides. So x = 64^(1/2) = 8. Check: 8^2 = 64.

Here log base x of 537824 = 5. Converting to exponential form: log_x(a) = k means x^5 = a. So x^5 = 537824. To solve for x, take the 5th root of both sides. So x = 537824^(1/5) = 14. Check: 14^5 = 537824.

Here log base x of 100000 = 5. Converting to exponential form: log_x(a) = k means x^5 = a. So x^5 = 100000. To solve for x, take the 5th root of both sides. So x = 100000^(1/5) = 10. Check: 10^5 = 100000.

Here log base 4 of x = 3. Converting to exponential form: log_a(x) = k means x = a^k. So x = 4^3. Computing 4^3 gives x = 64. Check: log base 4 of 64 = 3, since 4^3 = 64.

Here log base 6 of x = -2. Converting to exponential form: log_a(x) = k means x = a^k. Since k is negative, x = 6^-2 = 1/6^2. Now 6^2 = 36. So x = 1/36. Check: log base 6 of 1/36 = -2, since 6^2 = 36.

Here log base 3 of x = 2. Converting to exponential form: log_a(x) = k means x = a^k. So x = 3^2. Computing 3^2 gives x = 9. Check: log base 3 of 9 = 2, since 3^2 = 9.

Here log base 10 of x = 1. Converting to exponential form: log_a(x) = k means x = a^k. So x = 10^1. Computing 10^1 gives x = 10. Check: log base 10 of 10 = 1, since 10^1 = 10.

Here log base 11 of x = -4. Converting to exponential form: log_a(x) = k means x = a^k. Since k is negative, x = 11^-4 = 1/11^4. Now 11^4 = 14641. So x = 1/14641. Check: log base 11 of 1/14641 = -4, since 11^4 = 14641.

Here log base 3 of x = -3. Converting to exponential form: log_a(x) = k means x = a^k. Since k is negative, x = 3^-3 = 1/3^3. Now 3^3 = 27. So x = 1/27. Check: log base 3 of 1/27 = -3, since 3^3 = 27.

Here log base 3 of x = -4. Converting to exponential form: log_a(x) = k means x = a^k. Since k is negative, x = 3^-4 = 1/3^4. Now 3^4 = 81. So x = 1/81. Check: log base 3 of 1/81 = -4, since 3^4 = 81.