**Logarithmic and Exponential Functions Self Quiz--Solutions**

**1.** Compute:

**a. **log_{6} 216

**b. **

__Solutions__** **

a.log_{6}216 =log_{6}6^{3}=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__

*If you had difficulty with this exercise, review the tutorial Properties of Logarithms.*

**4.** Solve for *x*: 2^{3x+1}= 5^{x+6}

__Solution__

2^{3x+1}=5^{x+6}

*ln(*2^{3x+1})=*ln (*5^{x+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)=4 ^{x}* ?

__Solution__

When *x=0*, *f(x)=4 ^{0}=1. *Thus the

*y*-intercept of the graph of

*f(x)=4*is

^{x}*(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 2^{3}=3log 2=3(.301)=.903

** **

** b.** log 20= log((2^{2})(5))=log 2^{2} +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)=log _{5}x* ? The

*x*-intercept of the graph?

__Solution__

The function *g(x)=log _{5}x* is undefined at

*x=0*. The graph has no

*y*-intercept. However, when

*log*thus the graph of the function has an

_{5}x=0, x=5^{0}=1;*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, *log _{10}5* or

*ln 5*? Why?

__Solution__

Let *log _{10}5=a* and let

*ln 5=b*.

By definition, *10 ^{a}=5 *and

*e*Since

^{b}=5.*5<10*,

*a<1*; since

*e<5*,

*b>1.*

Thus *log _{10} 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))=2 ^{2}(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.*