Logarithms Review and Practice Questions

Logarithm questions appear on College Level math tests such as the Accuplacer and Compass.

A Logarithm is a quantity representing the power to which a fixed number (the base) must be raised to produce a given number.

Quick Review     Practice Questions   Answer Key

Logarithms – a Quick Review

Logarithm is a function that has the form
log_y x = a
It actually solves this equation: which number do we put as a degree on the variable y to get the variable x, that is:
y ^a = x
y is called the base and a is the exponent.
For example, let’s solve logarithm log₅25 = a.

                        5^a = 25
                        5^a = 5²

Here, we represent 25 using 5 and the second degree. a and 2 are both on the number 5, so they must be the same.

                        a = 2

We can see from the way the logarithm works, that:

log_a1 = 0 and log_aa = 1

From log_a1 = 0 we have that a⁰ = 1, which is true for any real number a.
From log_aa = 1 we have that a¹ = a, which is true for any real number a.

If in the logarithm the base is 10, then instead of log₁₀ we write lg.

When we are solving some logarithm, any part can be unknown. In the first example, we had a case where the exponent was the unknown variable. Let’s see another example, where both exponent and base are known:

lg x = 2
10² = x
                        x = 100

 

Practice Questions

1. If log₂x=3, then x is:

A. 9
B. 8
C. 7
D. 6

2. Solve the equation log₄1/4=x.

A. -1
B. 0
C. 1
D. 2

3. For what x is the following equation correct:

                        log_x125=3

A. 1
B. 2
C. 3
D. 5

4. Find x if log_x(9/25)=2.

A. 3/5
B. 5/3
C. 6/5
D. 5/6

5. Solve log₁₀10,000=x.

A. 2
B. 4
C. 3
D. 6

6. Find x if log_1/2 x=4.

A.16
B. 8
C. 1/8
D. 1/16

Answer Key

1. B. 8

log₂x=3
2³=x
x=8

2. A. -1

log₄1/4=x.
4^x=1/4
4^x=4^-1
x=-1

3. E. 5

log_x125=3
x³=125
x³=5³
x=5

4. A. 3/5

log_x(9/25) = 2
x²=9/25
x²=(3/5)²
x=3/5

5. B. 4

log₁₀10,000=x
10^x=10,000
10^x=10⁴
x=4

6. D. 1/16

log_1/2 x=4
(1/2)⁴=x
x=1/16

 

Written by: Brian Stocker MA, Complete Test Preparation Inc.
Modified: December 18th, 2018

Published: June 20th, 2014