Logarithms: IPMAT practice questions

Solution

The log is defined only for x − 4 > 0, that is x > 4.

The base 12 is less than 1, so the inequality reverses: x − 4 < (12)⁻³ = 8.

So 4 < x < 12, which holds for 7 integers (5 to 11).

Solution

log 7⁵³ = 53 × log 7 = 53 × 0.8451 = 44.7903.

A number whose log has integer part k has k + 1 digits.

The integer part is 44, so 7⁵³ has 45 digits.

Solution

log_a⁵ (b⁴) = (45) log_a b.

So it equals (45)k.

Solution

Both logs must be defined, so x > 8.

log₃ [x(x − 8)] = 2, so x² − 8x = 9.

x² − 8x − 9 = 0 gives (x − 9)(x + 1) = 0.

x = −1 is rejected (x must exceed 8), so x = 9.

Solution

log_a³ (b⁵) = (53) log_a b.

So it equals (53)k.

Solution

16 = 4², so log₄ 16 = 2.

Similarly the other two logs are 2 and 3.

Value = 2 + 2 − 3 = 1.

Solution

25 = 5², so 25^{log₅ 3} = 5^{2 log₅ 3} = (5^{log₅ 3})².

Since a^{log_a b} = b, this is 3² = 9.

Solution

216 = 6³, so log₆ 216 = 3.

Similarly the other two logs are 3 and 3.

Value = 3 − 3 + 3 = 3.

Solution

log 2⁵³ = 53 × log 2 = 53 × 0.3010 = 15.9530.

A number whose log has integer part k has k + 1 digits.

The integer part is 15, so 2⁵³ has 16 digits.

Solution

log_x 243 = 53 means x⁵³ = 243 = 3⁵.

Raise both sides to the power 35: x = 3³ = 27.

Solution

Write each log with a common base: log_p q = log q / log p.

The product is (log 6 / log 2) × (log 7 / log 6) × (log 16 / log 7) = log 16 / log 2.

That is log₂ 16 = 4.

Solution

4 = 2², so 4^{log₂ 4} = 2^{2 log₂ 4} = (2^{log₂ 4})².

Since a^{log_a b} = b, this is 4² = 16.

Solution

The log is defined only for x − 3 > 0, that is x > 3.

The base 12 is less than 1, so the inequality reverses: x − 3 < (12)⁻⁴ = 16.

So 3 < x < 19, which holds for 15 integers (4 to 18).

Solution

log_x 3 = 13 means x¹³ = 3 = 3¹.

Raise both sides to the power 31: x = 3³ = 27.

Solution

Both logs must be defined, so x > 26.

log₃ [x(x − 26)] = 3, so x² − 26x = 27.

x² − 26x − 27 = 0 gives (x − 27)(x + 1) = 0.

x = −1 is rejected (x must exceed 26), so x = 27.

Solution

9 = 3², so log₃ 9 = 2.

Similarly the other two logs are 3 and 2.

Value = 2 − 3 + 2 = 1.

Solution

log 2⁴⁴ = 44 × log 2 = 44 × 0.3010 = 13.2440.

A number whose log has integer part k has k + 1 digits.

The integer part is 13, so 2⁴⁴ has 14 digits.

Solution

Both logs must be defined, so x > 18.

log₃ [x(x − 18)] = 5, so x² − 18x = 243.

x² − 18x − 243 = 0 gives (x − 27)(x + 9) = 0.

x = −9 is rejected (x must exceed 18), so x = 27.

Solution

The log is defined only for x − 4 > 0, that is x > 4.

The base 12 is less than 1, so the inequality reverses: x − 4 < (12)⁻² = 4.

So 4 < x < 8, which holds for 3 integers (5 to 7).

Solution

Write each log with a common base: log_p q = log q / log p.

The product is (log 6 / log 3) × (log 13 / log 6) × (log 27 / log 13) = log 27 / log 3.

That is log₃ 27 = 3.

Solution

Write each log with a common base: log_p q = log q / log p.

The product is (log 5 / log 3) × (log 7 / log 5) × (log 9 / log 7) = log 9 / log 3.

That is log₃ 9 = 2.

Solution

log_a⁴ (b²) = (24) log_a b.

So it equals (12)k.

Solution

The sum of logs is the log of the product: log [(21)(32)(43)…(109)].

The product telescopes to 101 = 10.

log 10 = 1.

Solution

The sum of logs is the log of the product: log [(21)(32)(43)…(10099)].

The product telescopes to 1001 = 100.

log 100 = 2.