Review: Exponential and Logarithmic Functions
Basic skills Quiz 2 is over exponential and logarithmic functions. You need to know the
following:
Solving Equations Using Logs and Exponents
The natural log and the exponential functions are inverses of each other:
ln(e x ) = x
and
e ln(x ) = x
(The above is true wherever these are defined. Remember that the function ln(x) is only
defined when x>0.) We can use these properties to solve equations involving exponents
or logarithms:
•
If e 3t +1 = 4 , then ln(e 3t +1 ) = ln(4) . This becomes 3t + 1 = ln(4) , or t =
•
If ln(z) = 6y , then e ln(z ) = e 6y , or z = e 6y .
ln(4) −1
.
3
Be aware that you must apply the function to the entire side of the equation. For
example, if we have
6e 4 y = 3
and wish to solve for y, we should first divide both sides by 6 so we are left with just an
exponential on the left side:
e4y =
1
2
Then we can take the log of both sides:
4 y = ln( 12 )
or
y = 14 ln( 12 ) .
Exponential Properties
Don’t forget that the exponential function can be simplified using all the properties of
exponents you already know:
1. e a e b = e a +b
ea
2. b = e a−b
e
3.
(e )
a b
= ea b
We can use these to simplify expressions or solve equations:
e4z
= y , then e 4 z−z = e 3z = y , and z = 13 ln(y) .
ez
•
If
•
e 4 e t = e 4 +t .
Logarithm Properties
The logarithm function behaves in ways similar to exponents:
1. The log of a product is equal to the sum of the logs: ln(ab) = ln(a) + ln(b)
 a
2. The log of a quotient is equal to the difference of the logs: ln  = ln(a) − ln(b)
b
3. Taking the log of a power is the same multiplying: ln(a b ) = bln(a)
These properties are also useful:
•
If ln(x) + ln(4) = y , then we have ln(4 x) = y or x = 14 e y .
•
e ln(3)−ln(2) = e ln( 3 / 2) = 32 .
Converting Bases
We can use the third property of exponential functions listed above to convert bases. For
example, suppose we have f (x) = 2 x , and we wish to rewrite this with the base e. We
simply note that e ln(2) = 2 (since the logarithm is the inverse of the exponential function).
Then we write:
2 x = (e ln( 2) ) = e ln( 2)x .
x
We can then use the fact that ln(2) ≈ 0.6931 to write f (x) = 2 x ≈ e 0.6931x . (It is worth
noting that the last is only an approximation to the function, since ln(2) is not exactly
0.6931. However, it is a pretty good approximation.)
Examples:
•
e 6u = (e 6 ) ≈ 403.4 u since e 6 ≈ 403.4 .
•
7.4 r = (e ln( 7.4 ) ) ≈ e 2.001r since ln(7.4) ≈ 2.001 .
u
r
Practice Problems
Here are a few practice problems to help you prepare. Answers are on the next page.
1. If w = e
21x
4
, then ln(w)=
2. Rewrite e1.25x in the form a x .
3. Rewrite 0.6 y in the form e k y .
4. Simplify and rewrite e(ln(r )−ln( x)) without using ln or e in the final expression.
 e4 x 
5. Simplify and rewrite ln 5x  without using ln or e in the final expression.
e 
6. If ln(y) = 7x , then y=
Answers:
1.
21x
4
2. 3.4903x (since e1.25 ≈ 3.4903 .)
3. e−0.5108y (since ln(0.6) ≈ −0.5108 . Note that it is legal for the output of the
logarithm to be negative, just not the input.)
4. e(ln(r )−ln( x)) = e ln( r / x ) =
r
x
 e4 x 
5. ln 5x  = ln(e 4 x−5x ) = ln(e−x ) = −x
e 
6. y = e 7x