Logarithmic and Exponential Functions Self Quiz--Solutions
1. Compute:
a. log6 216
b. ![]()
Solutions
a. log6 216 =log6 63=3
b.
=
=
= -2
If you had difficulty with this exercise, review the tutorial Logarithmic Functions.
2. Solve for x: 
Solution


If you had difficulty with this exercise, review the tutorial Properties of Exponents.
3. Write as a single logarithm:
Solution
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If you had difficulty with this exercise, review the tutorial Properties of Logarithms.
4. Solve for x: 23x+1= 5x+6
Solution
23x+1=5x+6
ln(23x+1)=ln (5x+6)
(3x+1)ln(2)=(x+6)ln(5)
x(3ln(2)-ln(5))=6ln(5)-ln(2)
![]()
If you had difficulty with this exercise, review the tutorials Solving Exponential Equations and Solving Equations with Logarithms.
5. Write as a single exponential expression: 
Solution

If you had difficulty with this exercise, review the tutorial Properties of Exponents.
6. What is the y-intercept of the graph of f(x)=4x ?
Solution
When x=0, f(x)=40=1. Thus the y-intercept of the graph of f(x)=4x is (0,1).
If you had difficulty with this exercise, review the tutorial The Graph of an Exponential Function.
7. Given that log 2=.301 and log 5=.699, find
a. log 8
b. log 20
c. log .2
Solution
a. log 8=log 23=3log 2=3(.301)=.903
b. log 20= log((22)(5))=log 22 +log 5=2log 2+ log 5=2(.301)+.699=1.301
c. log .2=log 5-1 = -1(log 5)= -.699
If you had difficulty with this exercise, review the tutorial Properties of Logarithms.
8. What is the y-intercept of the graph of g(x)=log5x ? The x-intercept of the graph?
Solution
The function g(x)=log5x is undefined at x=0 . The graph has no y-intercept. However, when log5x=0, x=50=1; thus the graph of the function has an x-intercept at (1,0).
If you had difficulty with this exercise, review the tutorial The Graph of a Logarithmic Function.
9. Which is larger, log105 or ln 5? Why?
Solution
Let log105=a and let ln 5=b.
By definition, 10a=5 and eb=5. Since 5<10, a<1; since e<5, b>1.
Thus log10 5=a<1<b=ln 5.
If you had difficulty with this exercise, review the tutorial Two Common Bases For Logarithms.
10. The population of a colony of bacteria doubles every 5 hours. If 1000 bacteria
are present initially, how many bacteria does the colony contain after 2
days?
Solution
At time t=0 hours, the colony contains 1000 bacteria. After t=5 hours, the colony contains 2(1000)=2000 bacteria. After t=10 hours, it doubles again; now the colony numbers 2(2(1000))=22(1000) bacteria. Continuing, we see that in 2 days (or 48 hours), the population of the colony will have doubled
times; thus after 2 days, there are
bacteria in the colony.
If you had difficulty with this exercise, review the tutorial Real-World Applications of Exponential Functions.



= -2