LogarithmPracticeTest
IB Mathematics SL
1.
1
 – 2 in exponential form.
16
1
log 4    2
 16 
Rewrite the logarithmic equation log 4
4
1
log 4  
 16 
 42
1
 42
16
2.
1
in logarithmic form.
Rewrite the exponential equation 5–2 
25
1
5–2 
25
 1 
log5 52  log5  
 25 
 1 
2  log5  
 25 
 
3.
1
.
3
The bases of both the log and the exponential are the same and therefore are inverses of
each other.
1
 f( )
3
1
 log3  
3
Evaluate the function f ( x)  log3 x at x 
 
 log3 31
 1
4. Solve the equation below for x.
log 2 26  x
The bases of both the log and the exponential are the same and therefore are inverses of
each other.
log 2  26   x
6x
Logarithm Practice Test
5. Use the properties of logarithms to expand the expression.
1
log 2 16
5
The argument can be written as a power of 2.
1
 log 2 16
5
1
 log 2  24 
5
 4 1 
 log 2  2 5 


4
 
 log 2  2 5 
 
4

5
6. Determine the asymptote of y  log3  x  4  .
Exponential functions have a horizontal asymptote. Since log functions are inverse
funtions of exponentials, and since graphs of inverse functions reflect across the line
y  x , log functions must have a vertical asymptote. The parent function’s asymptote is
along the y-axis. This function is translated 4 units to the left and therefore, must be at
x  4 . Further evidence can be found by graphing the function. Note the use of the
change of base when defining Y1.
Page 2
Logarithm Practice Test
7. Use the properties of natural logarithms to rewrite the expression.
e ln 3.4
These two functions (by virtue of their common bases) are inverse functions of each
other.
 eln 3.4
 3.4
8. Use the properties of natural logarithms to rewrite the expression.
4 ln e0
Once again, since the two functions are inverses of each other, the problem is easily
simplified:
 4 ln  e0 
 40
0
9. Rewrite the logarithm log3 33 in terms of the natural logarithm.
This question is testing you on the Change-Of-Base property which states:
log c  b 
log a  b  
log c  a 
 log3 33

ln  33
ln  3
10. Identify the x-intercept of the function f ( x)  4 ln( x  2) .
0  4 ln( x  2)
0 4 ln  x  2 

4
4
0  ln  x  2 
ln x 2
e0  e  
1 x2
3 x
Page 3
Logarithm Practice Test
11. Evaluate the logarithm log 7 126 using the change of base formula. Round to 3 decimal
places.
The choice of base to use is arbitrary. Choose a base that your calculator uses like
log10 or ln . More information can be found on page 125 of your textbook.
 log 7 126

ln 126 
ln  7 
 2.485
12. Use the properties of logarithms to expand the expression as a sum, difference, and/or
constant multiple of logarithms. (Assume all variables are positive.)
y
7
For more information, consult page 116 of your book.
y
 log 3
7
 log 3  y   log 3  7 
log 3
13. Use the properties of logarithms to expand the expression as a sum, difference, and/or
constant multiple of logarithms. (Assume all variables are positive.)
log 3 x 2 y 5 z
For more information, consult page 116 of your book.
 log 3  x 2 y 5 z 
 log 3  x 2   log 3  y 5   log 3  z 
 2 log 3  x   5log 3  y   log 3  z 
Page 4
Logarithm Practice Test
14. Condense the expression log 7 x  log 7 3 to the logarithm of a single term.
For more information, consult page 116 of your book.
 log 7  x   log 7  3
 log 7  3x 
15. Condense the expression below to the logarithm of a single quantity.
6 ln x + 4 ln y – 5ln z
For more information, consult page 116 of your book.
 6 ln  x  + 4 ln  y  – 5ln  z 
 ln  x 6   ln  y 4   ln  z 5 
 ln  x 6 y 4 z 5 
 x6 y 4 
 ln  5 
 z 
16. Find the exact value of log 3 25 .
5
log5 3 25
1
log5  25 3 


1

3 
log5  52 


2
 log5  5 3 
 
2

3
 
Page 5
Logarithm Practice Test
17. Solve for x: 5 x /3  0.0052 . Round to 3 decimal places.
Although it would be possible to take log 5 of both sides of the equation, eventually, we
will have to use our calculator. It is therefore, wise to use one of the log functions that
our calculators are equipped to interpret such as log base-10 or the natural log. The
choice is yours as the results will be the same no matter which you choose to use.

5 x /3  0.0052

ln 5 x /3  ln  0.0052 
 x
    ln  5    ln  0.0052 
 3
ln  0.0052 
1
 x 
3
ln  5 
x
3  ln  0.0052  
ln  5 
x  9.803
18. Solve for x: e x (8  e x )  16 . Round to 3 decimal places.
Behold! The quadratic makes its appearance! Factor and solve.
e x (8  e x )  16
   
2
8 e x  e x  16
   8  e x   16
0   e x  4  e x  4 
0  ex
2
0  ex  4
4  ex
 
ln  4   ln e x
ln  4   x
1.386  x
Page 6
Logarithm Practice Test
19. Approximate the solution to ln 4 x  3.2 . Round to 3 decimal places.
To remove the natural log function we write both sides of the equation with a base of e.
This allows the variable x to be isolated.
ln  4 x   3.2
ln 4 x
e    e3.2
4 x  e3.2
e3.2
4
x  6.133
x
Page 7
Logarithm Practice Test
Answer Key
1.
1
16
2.
1
log5
–2
25
3. –1
4. 6
5. 4
5
6. x = -4
7. 3.4
8. 0
9. ln 33
ln 3
10. x = 3
11. 2.485
12. log 3 y  log 3 7
13. 2 log 3 x + 5log 3 y + log 3 z
14. log 7 3x
15.
x6 y 4
ln 5
z
16. 2
3
17. 9.803
18. 1.386
19. 6.133
4 –2 
Page 8