Logarithms: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 1)

The change-of-base formula rewrites log_a(x) using a different base b. It states log_a(x) = log_b(x) / log_b(a). Here b is any valid base, with b > 0 and b not equal to 1. This is useful when a calculator only computes base-10 or base-e logs. For example, log_2(8) = log(8) / log(2) = 3.

For a number greater than 1, count the digits to the left of the decimal point. The characteristic equals that digit count minus one. For example, 4567 has 4 digits before the decimal point. So its characteristic is 4 – 1 = 3. Check: log(4567) is about 3.66, with characteristic 3, matching the rule.

A common logarithm has two parts: a whole-number part and a decimal part. The decimal part is called the mantissa. The whole-number part is called the characteristic. For example, if log(x) = 2.531, the mantissa is 0.531. The mantissa is usually found using log tables.

A common logarithm has two parts: a whole-number part and a decimal part. The whole-number part is called the characteristic. The decimal part is called the mantissa. For example, if log(x) = 2.531, the characteristic is 2. The mantissa there is 0.531.

Logarithms with base 10 are called common logarithms. They are written log(x) or log₁₀(x), without stating the base. This is distinct from natural logarithms, which use base e. Natural logarithms are written ln(x) instead.

When a logarithm is written without a base, like log(x), the base is taken as 10. This is a fixed convention. A base-10 logarithm is called the common logarithm. So log(x) means log₁₀(x), unless another base is stated. A base-e logarithm is instead written ln(x).

Here we need log_a(a), for any valid base a (a > 0, a not equal to 1). Convert to exponential form: log_a(a) = m means a^m = a. Since a^1 = a, we get m = 1. So log_a(a) = 1. Check: a^1 = a, confirming the result.

Here we need log_a(1), for any valid base a (a > 0, a not equal to 1). Convert to exponential form: log_a(1) = m means a^m = 1. Any nonzero number raised to the power 0 equals 1. So a^0 = 1 for every valid base a. That means m = 0. So log_a(1) = 0. Check: a^0 = 1, confirming the result.

The quotient rule states log_a(x/y) = log_a(x) – log_a(y). Here x and y are positive numbers, with base a positive and a not equal to 1. This law turns a log of a quotient into a difference of logs. So log_a(x/y) equals log_a(x) minus log_a(y).

The product rule states log_a(xy) = log_a(x) + log_a(y). Here x and y are any positive numbers, with base a positive and a not equal to 1. This law turns a log of a product into a sum of logs. For example, log_a(x) + log_a(y) recombines into log_a(xy), and the reverse also holds.

Here a^m = x, with a positive and a not equal to 1. This is the defining relation for logarithms. By definition, m is written log_a(x). So m is called the logarithm of x to the base a. It is simply the exponent, rewritten as a log statement.

Here log base 625 of p = 2, so p = 625^2. Since 625 = 5^4, the law (a^m)^n = a^(mn) gives p = (5^4)^2 = 5^8. Also log base 5 of q = 4, so q = 5^4 = 5^4. Both p and q are powers of 5, with exponents 8 and 4. To relate them, raise each to make the total exponent match: p^1 has exponent 8 x 1 = 8. q^2 has exponent 4 x 2 = 8. These are equal, so p^1 = q^2.

Here log base 27 of p = 8, so p = 27^8. Since 27 = 3^3, the law (a^m)^n = a^(mn) gives p = (3^3)^8 = 3^24. Also log base 3 of q = 8, so q = 3^8 = 3^8. Both p and q are powers of 3, with exponents 24 and 8. To relate them, raise each to make the total exponent match: p^1 has exponent 24 x 1 = 24. q^3 has exponent 8 x 3 = 24. These are equal, so p^1 = q^3.

Here log base 4 of p = 2, so p = 4^2. Since 4 = 2^2, the law (a^m)^n = a^(mn) gives p = (2^2)^2 = 2^4. Also log base 2 of q = 8, so q = 2^8 = 2^8. Both p and q are powers of 2, with exponents 4 and 8. To relate them, raise each to make the total exponent match: p^2 has exponent 4 x 2 = 8. q^1 has exponent 8 x 1 = 8. These are equal, so p^2 = q^1.

Here log base 625 of p = 6, so p = 625^6. Since 625 = 5^4, the law (a^m)^n = a^(mn) gives p = (5^4)^6 = 5^24. Also log base 5 of q = 6, so q = 5^6 = 5^6. Both p and q are powers of 5, with exponents 24 and 6. To relate them, raise each to make the total exponent match: p^1 has exponent 24 x 1 = 24. q^4 has exponent 6 x 4 = 24. These are equal, so p^1 = q^4.

Here log base 27 of p = 5, so p = 27^5. Since 27 = 3^3, the law (a^m)^n = a^(mn) gives p = (3^3)^5 = 3^15. Also log base 3 of q = 5, so q = 3^5 = 3^5. Both p and q are powers of 3, with exponents 15 and 5. To relate them, raise each to make the total exponent match: p^1 has exponent 15 x 1 = 15. q^3 has exponent 5 x 3 = 15. These are equal, so p^1 = q^3.

Here log base 16 of p = 8, so p = 16^8. Since 16 = 2^4, the law (a^m)^n = a^(mn) gives p = (2^4)^8 = 2^32. Also log base 2 of q = 8, so q = 2^8 = 2^8. Both p and q are powers of 2, with exponents 32 and 8. To relate them, raise each to make the total exponent match: p^1 has exponent 32 x 1 = 32. q^4 has exponent 8 x 4 = 32. These are equal, so p^1 = q^4.

Here log base 16 of p = 4, so p = 16^4. Since 16 = 2^4, the law (a^m)^n = a^(mn) gives p = (2^4)^4 = 2^16. Also log base 2 of q = 4, so q = 2^4 = 2^4. Both p and q are powers of 2, with exponents 16 and 4. To relate them, raise each to make the total exponent match: p^1 has exponent 16 x 1 = 16. q^4 has exponent 4 x 4 = 16. These are equal, so p^1 = q^4.

Here log base 25 of p = 3, so p = 25^3. Since 25 = 5^2, the law (a^m)^n = a^(mn) gives p = (5^2)^3 = 5^6. Also log base 5 of q = 3, so q = 5^3 = 5^3. Both p and q are powers of 5, with exponents 6 and 3. To relate them, raise each to make the total exponent match: p^1 has exponent 6 x 1 = 6. q^2 has exponent 3 x 2 = 6. These are equal, so p^1 = q^2.

Here we need log base 5 of (25)^4. The power rule log_a(x^p) = p x log_a(x) applies here. First find log base 5 of 25. Since 25 = 5^2, log base 5 of 25 = 2. Now apply the power rule: 4 x 2 = 8. So log base 5 of (25)^4 = 8. Check: 5^8 = (5^2)^4 = 25^4 = 390625.

Here we need log base 3 of (27)^3. The power rule log_a(x^p) = p x log_a(x) applies here. First find log base 3 of 27. Since 27 = 3^3, log base 3 of 27 = 3. Now apply the power rule: 3 x 3 = 9. So log base 3 of (27)^3 = 9. Check: 3^9 = (3^3)^3 = 27^3 = 19683.

Here we need log base 2 of (32)^2. The power rule log_a(x^p) = p x log_a(x) applies here. First find log base 2 of 32. Since 32 = 2^5, log base 2 of 32 = 5. Now apply the power rule: 2 x 5 = 10. So log base 2 of (32)^2 = 10. Check: 2^10 = (2^5)^2 = 32^2 = 1024.

Here we need log base 6 of (36)^3. The power rule log_a(x^p) = p x log_a(x) applies here. First find log base 6 of 36. Since 36 = 6^2, log base 6 of 36 = 2. Now apply the power rule: 3 x 2 = 6. So log base 6 of (36)^3 = 6. Check: 6^6 = (6^2)^3 = 36^3 = 46656.

Here we need log base 5 of (625)^3. The power rule log_a(x^p) = p x log_a(x) applies here. First find log base 5 of 625. Since 625 = 5^4, log base 5 of 625 = 4. Now apply the power rule: 3 x 4 = 12. So log base 5 of (625)^3 = 12. Check: 5^12 = (5^4)^3 = 625^3 = 244140625.

Here we need log base 2 of (4)^4. 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: 4 x 2 = 8. So log base 2 of (4)^4 = 8. Check: 2^8 = (2^2)^4 = 4^4 = 256.