Exponential and Logarithmic Functions

Lesson 6 Eponential and Logarithmic Fu tions Lesson 6 Eponential and Logarithmic Functions Eponential functions are of the form y = a where a is a constant greater than zero and not equal to one and is a variable. Both y = and y = e are eponential functions. The function, e, is etensively used in calculus. You should memorize its approimate value when =. (e.78) You should also be able to quickly graph y = e without the aid of a calculator. A simple graph of y = e is shown below. Figure 3 3 y = e Logarithmic FunctionS The equation y = log a is the same as a y =. The inverse of the eponential function is y = a. In this course we will restrict our study of logarithms to log base e which will be written as ln(). The equation y = ln() is the inverse function of y = e. Notice that the graph of ln() is a reflection of graph of e around the line y =. You should be able to quickly sketch from memory y = ln(). It will also be important to remember the basic logarithm rules listed on the net page. EXPONENTIAL AND LOGARITHMIC FUNCTIONS - LESSON 6 59

Figure y = ln() I. ln() = 0 II. ln(e) = III. IV. ln(e ) = e ln() = V. Product: ln(y) = ln() + ln(y) VI. Quotient: ln(/y) = ln() ln(y) VII. Power: ln( a ) = a ln() Remember that the natural log of a negative number is undefined. Some books specify ln() as ln. We will use ln() for this book. Be careful to use only positive, non-zero values for when employing the natural log function. 60 LESSON 6 - EXPONENTIAL AND LOGARITHMIC FUNCTIONS CALCULUS

The natural log function can be used to free variable eponents from their eponential functions. Conversely, the eponential function can do the same for the natural log functions. Eample Solve for. e = Taking ln of both sides: ( ) = () ln e ln = 0 = 0 checking e (0) = e 0 = Eample Solve for. ln( + 5) = 0 Use each side of the equation as the eponent for e. e ln( + 5) = e 0 + 5 = ; so = 4 Sometimes the equations are comple and we need to use substitution to solve them. See eample 3 on the net page. CALCULUS EXPONENTIAL AND LOGARITHMIC FUNCTIONS - LESSON 6 6

Eample 3 Solve for. e 4e + 3 = 0 Substituting u = e, u 4u + 3 = 0. Factoring, we get (u 3)(u ) = 0. Replacing u with e, we get (e 3)(e ) = 0. Solving each factor, we get: e = 3; e =. Taking the ln of both sides: ( ) = ( ) ln e ln 3 = ( ) ln 3 ( ) = () ln e ln = 0 Eample 4 Draw the graph of y = e and its inverse. ln ln y = e y = e switch variables y = e ( ) ( ) ( ) = f = ( y) ln e = y ln ( ) y = e y = f ( ) = ln ( ) 6 LESSON 6 - EXPONENTIAL AND LOGARITHMIC FUNCTIONS CALCULUS

l e s s o n p r a c t i c e 6A Answer the question.. Draw the graph of y = e 3. Find the inverse function. Graph it.. Draw the graph of y = e. Find the inverse function. Graph it. 3. Solve for. A. e + = B. e 3 = e 0 C. 0 = ln( +5) calculus Lesson Practice 6A 45

LESSON PRACTICe 6A D. ln() + ln(5) = 6 4. Solve for. ( Hint: Substitute and factor.) A. e 5e = 6 B. e + 7e = 4 46 calculus

l e s s o n p r a c t i c e 6B Answer the question.. + Draw the graph of y = e. Find the inverse function. Graph it.. Draw the graph of y = e. Find the inverse function. Graph it. 3. Solve for. A. e + ln(3) = B. e + = e calculus Lesson Practice 6B 47

LESSON PRACTICe 6B C. ln( + 3 + 5) = ln( ) D. ln ( ) = 3 4. Solve for. A. ln () + 3 = 7ln() B. e = e 48 calculus

l e s s o n p r a c t i c e 6C Answer the question.. Draw the graph of y =. Find the inverse function. Graph it. Solve for.. 4 e = e 3. ln(3 ) = calculus Lesson Practice 6C 49

LESSON PRACTICe 6C 4. e 7e + 0 = 0 5. ln () = ln() 6. e 3e + = 0 50 calculus

l e s s o n p r a c t i c e 6D Solve for.. + e = 5. e + 5e = 3 3. ln( + ) = calculus Lesson Practice 6D 5

LESSON PRACTICe 6D 4. ln( + ) + ln(4) = 3 5. Solve for : ln( 4) =. 6. 3 Draw the graph of y = e. Find the inverse function. Graph it. 5 calculus

t e s t 6 Circle your answer. ( ). Simplify ln 9 = A. ln(4.5) B. ln(4.5) C. ln(3) D. cannot be simplified. Solve for : ln() ln(4) =. A. 4 e B. e 4 C. e D. 4e 3. ln 6 () is the same as: 3 A. B. ln(8) C. ln(3) D. ln(6) ln(3) 4. Find the inverse function: f() = ln( ). A. f () = ln( + ) B. f () = e + C. f () = e D. f () = e 5. Simplify ln ( ) + ln ( 0 ). A. ln ( ) B. 0 ( ) C. ln 5 D. cannot be simplified calculus Test 6 7

Test 6 6. Solve for : ln () 5 ln() = 4 A. = e, e 4 B. = ln(4), ln(5) C. = e 5, e D. = ln(4), e 7. e X and ln() are inverse functions. The graph of y = ln() is the reflection of the graph of y = e around the: A. -ais B. y-ais C. origin D. line y = 8. Solve for : e = 3e A. = e 3 B. = ln(3) C. = ln(3) and = 0 D. = 0 9. Solve for : ln() + ln() = 7 A. = e 7 B. = 7e C. = e 7 D. = e7 0. Solve for : e 3 = A. 3 B. 3 C. 3 D. e 3 8 calculus

( ) 4 7. cos( ) = 0 for in [0, π] cos( θ) = 0 when θ = π, 3π, 5π etc = π = 3π = 5π = π = 3π = 5π 4 4 4 = π and 3π 4 4 [ ] 8. tan( ) = 0 0, π tan ( θ) = 0 when θ = 0, π, π, 3π etc. = 0 = π = 0 = π [ ] = 0 is the only answer in 0, π Lesson Practice 6A. y = e 3 y = e 3 y 3 = e ( ) = ( y ln 3 ln e ) ln( 3) = y f ( ) = ln( 3) f () = e 3 ( Switch variables) y = f () = ln(3). Lesson Practice 5D - LESSON PRACTICE 6A ln note: y = e y e reverse variables y ln( ) = ln( e ) = ( ) ln( ) = ln + y ( ) ln( ) = y ( ) = ( ) ( ) f ln ln This problem is the same as eample 4 in the instruction manual but solved differently. Both solutions are correct. ( ) ( ) = () ln ln ln y = e + 3. A. e = ln( + e ) = ln() + = 0 = = 3 0 B. e = e ( 3 e ) = ( 0 ln ln e ) ln( ) + 3 = 0 3 = ln( ) = ln 3 y = f ()= ln( ) ln() ( ) calculus solutions 69