Exponential Eqns - Intermediate Algebra - Lecture Slides, Slides of Algebra

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§9.5a

Exponential Eqns

Review §

 Any QUESTIONS About

• §9.4 → Logarithm Change-of-Base

 Any QUESTIONS About HomeWork

• §9.4 → HW-

9.4 MTH 55

Typical Log-Confusion

• Beware that Logs do NOT behave Algebraically. In General:

log log , log

a a a

M M

N N

log ( a MN ) ≠ (log (^) a M )(log (^) a N ),

log ( a M + N ) ≠ log (^) a M + log (^) a N ,

log ( a M − N ) ≠ log (^) a M − log (^) a N.

Solving Exponential Equations

• Equations with variables in exponents,

such as 3 x^ = 5 and

73 x^ = 90 are called

EXPONENTIAL EQUATIONS

• Certain exponential equations can be

solved by using the principle of

exponential equality

Example  Exponential Equality

• Solve for x : 5 x^ = 125
• SOLUTION
• Note that 125 = 5^3. Thus we can write each side as a power of the same base: 5 x^ = 5 3
• Since the base is the same, 5, the exponents must be equal. Thus, x must be 3. The solution is 3.

Example  Exponential Equality

• Solve each Exponential Equation

a. 25 x^ = 125 b. 9 x^ = 3 x +^1

 SOLUTION

a. (^) ( ) 5 2

x = 5 3 5 2 x^ = 5 3 2 x = 3

x =

b. (^) ( ) 32

x = 3 x +^1

32 x^ = 3 x +^1 2 x = x + 1 2 x − x = 1 x = 1

Example  Logarithmic Equality

• Solve for x : 3 x +1^ = 43
• SOLUTION 3 x^ +1^ = 43

log 3 x^ +1^ = log 43 ( x +1)log 3 = log 43 x +1 = log 43/log 3 x = (log 43/log 3) – 1 x ≈ 2.4236.

Principle of logarithmic equality Power rule for logs

 The solution is (log 43/log 3) − 1, or approximately 2.4236.

Example  Logarithmic Equality

• Solve for t : e 1.32 t^ = 2000
• SOLUTION

t ≈ 5.7583.

Note that we use the natural logarithm Logarithmic and exponential functions are inverses of each other

e 1.32 t^ = 2000

ln e 1.32 t^ = ln 2000

1.32 t = ln 2000

t = (ln 2000)/1.

Example  Solve by Taking Logs

• Solve each equation and approximate the results to three decimal places. a. 2 x^ = 15 b. 5 ⋅ 2 x −^2 = 17

 SOLUTION a. 2 x^ =^15

ln 2 x^ = ln

x ln 2 = ln

x =

ln

ln 2

Example  Solve by Taking Logs

• SOLUTION

b. 5 ⋅ 2 x −^3 = 17

2 x −^3 =

ln 2 x −^3 = ln

^

(^ x^ −^3 )ln 2^ =^ ln^

^

x − 3 =

ln

^

ln 2

x =

ln

^

ln 2

x ≈ 4.

b. 5 ⋅ 2 x −^2 = 17

Example  Eqn Quadratic in Form

• Solve for x : 3 x^ − 8∙3 −x^ = 2.
• SOLUTION

3 x^ ( 3 x^ − 8 ⋅ 3 −^ x )= 2 3 ( ) x

32 x^ − 8 ⋅ 30 = 2 ⋅ 3 x 32 x^ − 8 = 2 ⋅ 3 x 32 x^ − 2 ⋅ 3 x^ − 8 = 0  This equation is quadratic in form.  Let y = 3 x^ then y^2 = (3 x ) 2 = 3^2 x. Then,

Example  Eqn Quadratic in Form

• Soln cont.

32 x^ − 2 ⋅ 3 x^ − 8 = 0 y^2 − 2 y − 8 = 0 (^ y^ +^2 )( y^ −^4 ) =^0 y + 2 = 0 or y − 4 = 0 y = − 2 or y = 4 3 x^ = − 2 or 3 x^ = 4  But 3 x^ = −2 is not possible because 3 x^ > 0 for all numbers x. So, solve 3 x^ = 4 to find the solution

Example  Population Growth

• The following table shows the approximate population and annual growth rate of the United States of America and Pakistan in 2005

Country Population

Annual Population Growth Rate USA 295 million 1.0% Pakistan 162 million 3.1%

Example  Population Growth

• Use the population model P = P 0 (1 + r ) t^ and the information in the table, and assume that the growth rate for each country stays the same.
• In this model,
  • P 0 is the initial population,
  • r is the annual growth rate as a decimal
  • t is the time in years since 2005