Exponential and logarithmic functions (pp ) () Supplement October 14, / 1. a and b positive real numbers and x and y real numbers.

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1 MA123, Supplement Exponential and logaritmic functions pp ) Capter s Goal: Review te properties of exponential and logaritmic functions. Learn ow to differentiate exponential and logaritmic functions. Learn about exponential growt or decay penomena. ) Supplement October 14, / 1 Laws of Exponents a and b positive real numbers and x and y real numbers. a 0 = 1 a x = 1 a x a x a y = a x+y a x a y = a x y a x ) y = a xy ab) x = a x b x a b ) x = a x b x ) Supplement October 14, / 1

2 Exponential Functions a 1 positive real number. fx) = a x is called an exponential function wit base a. a > 1 : exponential growt blue curve) 0 < a < 1 : exponential decay red curve) ) Supplement October 14, / 1 Derivative of Exponential Function fx) = a x. Wat is f x)? so fx + ) fx) = ax+ a x ) = ax a a x a 1 = a x, f x) = lim 0 fx + ) fx) a 1 But wat is lim? 0 = lim 0 ) a 1 a x, ) Supplement October 14, / 1

3 A Fundamental Limit: Te number e = is base of te natural exponential function. Fill te table below e Tus e 1 lim 0 = 1 ) Supplement October 14, / 1 Derivatives of exponential functions: Now Teorem e x ) = lim 0 ) e 1 e x = 1 e x = e x. e x ) = e x. Verbally: Te natural exponential function is its own derivative. Moreover, applying te cain rule d dx egx) ) = e gx) d dx gx)) e gx) ) = e gx) g x) ) Supplement October 14, / 1

4 Te Natural Logaritm Te natural logaritm, ln x, is te inverse function to e x : e ln x = x and ln e x = x. Alternatively, y = e x if and only if x = ln y. Te grap of lnx) in red) is obtained by reflecting y = e x in blue) across y = x dotted). ) Supplement October 14, / 1 Properties of Logaritms A and B are positive real numbers, p any real number. ln 1 = 0 lna B) = ln A + ln B ln A ) = ln A ln B B lna p ) = p ln A Domain of y = ln x is 0, ). lim ln x = x 0 + lim ln x = x Please note: Tere is no particularly nice way to simplify lna + B). In particular lna + B) lna) + lnb). ) Supplement October 14, / 1

5 Derivatives of logaritmic functions: Wat is ln x)? Using e ln x = x we see e ln x ) = x = 1. On te oter and e ln x ) = e ln x ln x) = x ln x) So x ln x) = 1; solving for te unknown ln x) : ln x) = 1 x. ) Supplement October 14, / 1 Derivatives of logaritmic functions: Teorem ln x) = 1 x d In Leibniz notation: dx [ln x] = 1 x Moreover, applying te cain rule d 1 d [ln gx)] = dx gx) dx [gx)] ln gx)) = g x) gx). ) Supplement October 14, / 1

6 Wat about more general derivatives? Observe tat we ave te identities a x = e lnax) = e x ln a log a x = ln x ln a. Tus using te previous results we obtain te following formulas for te derivatives of general exponential and logaritmic functions d dx ax ) = a x ln a and d dx log a x) = 1 x ln a. Note: Let fx) = 3 x. In Example 8 of Capter 4, we saw tat f 1) Using te above formula we ave tat f x) = 3 x ln3), so tat te exact value for f 1) is 3 ln3) = ln3 3 ) = ln27). ) Supplement October 14, / 1 Wy does te Power Rule Work? We ve made eavy use of x n ) = nx n 1, but wy does tis work? Examples 6-12 in Capter 4 proves tis for a few values of n, but none of tose arguments seem capable of proving it for general n. Lets view x n ) as our unknown. Now so ln x n ) = n ln x ln x n )) = n ln x) = n x. On te oter and using cain rule) ln x n )) = xn ) x n Solving for our unknown x n ) : so x n ) x n = n x. x n ) = n x xn = nx n 1. ) Supplement October 14, / 1

7 Example 1: Find te derivative wit respect to x of fx) = e 4x. e 4x ) = 4x) e 4x = 4e 4x. Evaluate f x) at x = 1/4. f 1 4 ) = 4e4 1/4) = 4e. Compute f x), f x) and f 10) x). Can you guess wat te nt derivative f n) x) of fx) looks like? f x) = 4e 4x ) = 4 4e 4x = 4 2 e 4x. f x) = 4 2 e 4x ) = 4 2 4e 4x = 4 3 e 4x. Eac derivative multiplies e 4x by anoter factor of 4: f n) x) = 4 n e 4x. ) Supplement October 14, / 1 Example 2: Find te derivative wit respect to x of gx) = x 2 e x. Evaluate g x) at x = 1. Tis requires te product rule: x 2 e x ) = x 2 ) e x + x 2 e x ) = 2xe x + x 2 e x = 2 + x)xe x. So g 1) = 2 + 1) 1 e 1 = 3e. ) Supplement October 14, / 1

8 Example 3: Find te derivative wit respect to t of ft) = e 3t 4. Tis requires te cain rule: e 3t 4) = 3t 4 ) e 3t 4. But 3t 4 ) = 3t 4) 1/2 ) = 1 2 3t 4) 1/2 3t 4) = 1 2 3t 4) 1/2 3 = 3 3t 4) 1/2 2 So f t) = 3 2 3t 4) 1/2 e 3t 4 = 3e 3t 4 2 3t 4 ) Supplement October 14, / 1 Example 4: Find te derivative wit respect to x of y = lne x ). Te natural logaritm and e x are inverses so ln e x = x. Terefore lne x )) = x = 1. Suppose we forgot tat lne x ) = x. We could ave still found te derivative: ln e x ) = ex ) = ex e x e = 1. x ) Supplement October 14, / 1

9 Example 5: Find te derivative wit respect to x of fx) = x ln x. Tis requires te product rule: x ln x) = x ln x + x ln x) = ln x + x 1 x = ln x + 1. ) Supplement October 14, / 1 Example 6: Find te derivative wit respect to x of y = ln5x + 1). Tis requires te cain rule: Outside = ln ) Outside = 1 ) Inside = 5x + 1 Inside = 5. ln5x + 1)) = 1 5x = 5 5x + 1. ) Supplement October 14, / 1

10 Example 7: Find te derivative wit respect to x of gx) = ln3x 4 7x 2 + 5). Tis requires te cain rule: Outside = ln ) Outside = 1 ) Inside = 3x 4 7x Inside = 12x 3 14x. ln3x 4 7x 2 + 5) ) = 1 3x 4 7x x3 14x) = 12x3 14x 3x 4 7x ) Supplement October 14, / 1 Example 8: Find te derivative wit respect to x of fx) = lnlnln x)). But Tus lnlnx))) = f x) = lnlnx))) lnlnx)) f x) = ln x) ln x = 1/x ln x = 1 x ln x. 1 xln x)lnlnx))) fx) = lnlnln x)) as been proven to go to infinity, but nobody as observed it do so. Indeed, to get fx) > 1 we need x 4, 000, 000. To get fx) > 2 need x x needs to ave more tan 700 digits!) ) Supplement October 14, / 1

11 Example 9: Find te derivative wit respect to x of x) = e x2 +3 ln x. ) e x2 +3 ln x = x ln x) e x2 +3 ln x = 2x ) e x2 +3 ln x x = 2x + 3 ) e x2 +3 ln x. x ) Supplement October 14, / 1 Te Law of Natural Growt Let Qt) denote te quantity of some substance as a function of time t. In many applications te rate at wic a quantity grows or srinks varies directly wit current amount of te quantity. Symbolically, Q t) = rqt). Te larger a bacteria culture is, te faster it will grow Te more money in your bank account, te faster your account will grow Te larger a sample of radioactive material, te more frequently it will emit radiation. Teorem If Q t) = rqt) ten Qt) = Q 0 e rt for some number Q 0. If Q t) = rqt) ten Qt) = Q 0 e rt for some number Q 0. ) Supplement October 14, / 1

12 Example 10: Te grap of te function gx) passes troug te point 0, 5). It is known tat te slope of g at any point P is 7 times te y P -coordinate of P. Find g2). Te slope at a point is g x), te y coordinate of tat point is gx). Tus g x) = 7gx) for all x and so for some number C. But gx) = Ce 7x 5 = g0) = Ce 7 0 = C. So gx) = 5e 7x. ) Supplement October 14, / 1 Example 11: If $10, 000 is invested at 6% interest, find te value of te investment at te end of 8 years if te interest is compounded continuously. Let At) be te accumulated value after t years. Continuously compounded interest at 6% implies A t) = 0.06At). Applying te law of natural growt wit r = 0.06, But Tus, and At) = A 0 e 0.06t. 10, 000 = A0) = A 0 e = A 0. At) = 10, 000e 0.06t A8) = 10, 000e = 10, 000e 0.48 = ) Supplement October 14, / 1

13 Example 12: How many years will it take an investment to quadruple in value if te interest rate is 7% compounded continuously? Let A 0 denote te initial amount and At) be te accumulated value after t years. By te law of natural growt At) = A 0 e 0.07t. Now find t so tat At) = 4A 0 4A 0 = A 0 e 0.07t 4 = e 0.07t ln 4 = 0.07t so t = ln = 19.8 years ) Supplement October 14, / 1 Banker s Rule of 70 Optional) Bankers often use te following approximation to find te doubling time of an investment: Time to double 70 Interest rate as a percent). Te doubling time in te previous example is tus 70 7 = 10. To quadruple, te investment would ave to double twice, so te quadrupling time sould be twice te doubling time: Time to quadruple 2 10 = 20 years. ) Supplement October 14, / 1

14 Example 13: An amount of P 0 dollars is invested at 5% interest compounded continuously. Find P 0 if at te end of 10 years te value of te investment is $20, 000. Let At) denote te accumulated valued after t years. Te law of natural growt gives At) = P 0 e 0.05t. But A10) = 20, 000 so 20, 000 = P 0 e = P 0 e 0.5 Multiplying troug by e 0.5 gives P 0 = 20000e 0.5 = ) Supplement October 14, / 1 Example 14: A bacteria culture starts wit 2, 000 bacteria and te population triples after 5 ours. Find a and b, were number of bacteria after t ours satisfies yt) = a e bt 2000 = y0) = a e b 0 = a and so yt) = 2000e bt. Te bacteria triples in 5 ours so y5) = = = y5) = 2000e 5b 3 = e 5b so ln 3 = 5b, b = ln 3 5. ) Supplement October 14, / 1

15 Example 15: A bacteria culture starts wit 5, 000 bacteria and te population quadruples after 3 ours. Find an expression for te number P t) of bacteria after t ours. If P t) is te number of bacteria after t ours ten P t) = 5000e rt. But P 3) = = and so = 5000e 3r so 4 = e 3r so So ln 4 = 3r so r = ln 4 3. P t) = 5000e ln 4 3 t = 5000 e ln 4) t/3 = t/3. ) Supplement October 14, / 1 Example 16: If te bacteria in a culture doubles in 3 our, ow many ours will it take before 7 times te original number is present? First, roug estimate: Doubles after 3 ours, so 4 times original after 6 ours, and 8 times original after 9 ours. Time to reac 7 times original amount will be between 6 and 9 ours, and closer to 9 tan 6. Exact: Qt) = Q 0 e rt satisfies 2Q 0 = Q 0 e 3r. 2 = e 3r so r = ln 2 3 so Now find t so tat 7Q 0 = Qt). Qt) = Q 0 e ln2)/3) t. 7Q 0 = Q 0 e ln2)/3) t so 7 = e ln2)/3) t ln 7 = ln 2 3 t so t = 3 ln 2 ln 7 = ours ) Supplement October 14, / 1

16 Example 17: If te world population in 2010 will be 6 billion and it were to grow exponentially wit a growt constant r = 1 ln 2, find te 30 population in billions) in te year Let P t) denote world population in billions t years after P t) = 6e 1 30 ln 2)t. Population in 2070 is ten P 60) = 6e 1 30 ln 2) 60 = 6e 2 ln 2 = 6 4 = 24. Population would be 24 billion. ) Supplement October 14, / 1 Example 18: Te alf-life of Cesium-137 is 30 years. Suppose we ave a 100 gram sample. How muc of te sample will remain after 50 years? Qt) is te amount of Cesium-137 in grams t years from te present. But Qt) = 100e rt. 50 = Q30) = 100e 30r alf is left after 30 years) So 1 2 = e 30r. Multiply by 2e 30r : so e 30r = 2 and so r = ln ln 2 Qt) = 100e 30 ) t ln 2 Q50) = 100e 30 ) 50 = 100e = 31.5 grams. ) Supplement October 14, / 1

17 Example 18 continued): Roug estimate: After 30 years te initial 100 grams are cut in alf, to 50 grams. After an additional 30 years te 50 grams are cut in alf, to 25 grams. Tus, after 60 years tere are 25 grams left. Tus, after 50 years, te amount left will be between 25 and 50 grams, and sould be closer to 25 tan to 50. Our roug estimate is consistent wit te exact value of 31.5 grams we found. ) Supplement October 14, / 1