Extra Practice Exponentials and Logarithms

EXTRA PRACTICE: Exponentials and Logarithms 1. Rewrite using the rules of logarithms and exponents: 1 (a) 2log2 (x−5)−2 = 2log2 (x−5) 2−2 = (x − 5) 4 √ 2 ! 1 5x (b) 2 ln(x) − 3 ln(x − 2) + ln(5) = ln 2 (x − 2)3 (c) ln(3e−5x+2 ) = ln(3) + ln(e−5x+2 ) = ln(3) − 5x + 2 (d) ln(3 + e−5x+2 ) No simplification possible. 2 1 (e) − log5 (5m2 ) + log5 (25m2 ) 3 2 log5 ((5m2 )−2/3 ) + log5 (5m) = log5 (5−2/3 51 m−4/3 m1 ) = log5 (51/3 m−1/3 ) = 1 1 − log5 (m) 3 3 2. Write the exponential expression as an equivalent logarithm, or write the logarithm as an equivalent exponential. √ (a) log√3 (81) = 8 ⇒ ( 3)8 = 81 ⇒ 34 = 81 (b) (2/3)−3 = 27/8 ⇒ 2y = r ⇒ (c) log2 (r) = y ln(27/8) ln(2/3) log2/3 (27/8) = −3 or −3 = 3. Compute each of the following quantities on your calculator: (4 decimals is normally good enough) (a) log√3 (5) ⇒ ln(5) ≈ 2.9299 ln(3) 1 2 (b) 5log2 (3) Compute the exponent first: log2 (3) = imation is 12.8186 (c) log2 (56) = ln(56) ln(2) ln(3) ln(2) ≈ 1.58496, so the final approx- ≈ 5.8074 4. Solve for x: (a) ln(3x + 2) + ln(x − 1) = 1 ln((3x + 2)(x − 1)) = 1 ⇒ 1± (3x + 2)(x − 1) = e ⇒ 3x2 − x − (2 + e) = 0 q 1 + 12(2 + e) ≈ 1.432, −1.0984 6 We disregard the second solution, which would make x − 1 < 0 x= 1 (b) ln(x2 ) = (ln(x))2 (ln(x))2 − 2 ln(x) = 0 ⇒ ln(x) (ln(x) − 2) = 0 So either ln(x) = 0 (in which case x = e0 = 1) or ln(x) = 2 (in which case x = e2 ). (c) 32x−1 = 4x+2 . Take the log of both sides (any base will do, but we normally would choose ln):    ln 32x−1 = ln 4x+2  ⇒ ⇒ 2 ln(3)x−ln(3) = ln(4)x+2 ln(4) x= (d) (2x − 1) ln(3) = (x + 2) ln(4) ⇒ (2 ln(3)−ln(4))x = 2 ln(4)+ln(3) ⇒ 2 ln(4) + ln(3) ≈ 4.7738 2 ln(3) − ln(4) 4e2x =1 2 + e2x 4e2x = 2 + e2x 3e2x = 2 ⇒ e2x = ⇒ 2 3 ⇒ 1 2 ln 2 3   x= (e) ln(8x − 7) − ln(3 + 4x) = ln(9/11) 8x − 7 9 ln = ln 3 + 4x 11     8x − 7 9 = 3 + 4x 11 ⇒ ⇒ 52x = 104 ⇒ 11(8x−7) = 9(3+4x) x=2 5. Differentiate: (a) y = x2 + 4 − ln(3x) (b) y = 5x − log3 (x) (c) y = 23 x (d) y = 2log4 (x) 1 1 · 3 = 2x − 3x x 1 y 0 = 5x ln(5) − x ln(3) y 0 = 2x − x y 0 = 23 ln(2) · 3x ln(3) y 0 = 2log4 (x) ln(2) · 1 x ln(4) (e) y = 7tan(x) − log2 (x2 − 3x + 5) y 0 = 7tan(x) ln(7) · sec2 (x) − (f) y = 3arcsin(x) y 0 = 3arcsin(x) ln(3) · √ 2 (x2 1 · (2x − 3) − 3x + 5) ln(2) 1 1 − x2 ⇒ (g) y = log7 (x2 − 3) + log4 (x) 1+x2 1 y = 2 · 2x + (x − 3) ln(7) 0 1 (1 x ln(4) + x2 ) − log4 (x)(2x) (1 + x2 )2 (h) y = x3x . Here we need to rewrite: x3x = eln(x  y 0 = e3x ln(x) · 3 ln(x) + 3x 3 x) = e3x ln(x) . Now, 1 = x3x (3 ln(x) + 3) x  2x (i) y = π e 2x x y 0 = π e ln(π) · e2 · 2x ln(2) 2 (j) y = (1 + x2 )x Rewrite (1 + x2 )x = ex ln(1+x ) . 0 y =e x ln(1+x2 ) 1 2x2 2 x 2 · ln(1 + x ) + x · 2x = (1 + x ) ln(1 + x ) + 1 + x2 1 + x2   2 ! 6. Give a formula for differentiating each of the following (assume a > 0 is a constant): (a) af (x) This is a chain rule problem, af (x) ln(a) · f 0 (x) or f 0 (x) ln(a) · af (x) Here are examples of problems of this type: 2x (b) (f (x))a 2 −5x+1 3sin(x) π log3 (x) 32 x This is a chain rule problem, a(f (x))a−1 f 0 (x) or f 0 (x) · a · (f (x))a−1 Here are examples of problems of this type: (1 + x2 )5 (c) xf (x) Rewrite as eln(x e f (x)·ln(x) f (x) ) (log4 (x))e sinπ (x) = ef (x)·ln(x) . Now, 1 f (x) = xf (x) f 0 (x) ln(x) + · f (x) ln(x) + f (x) x x   0 (d) loga (f (x)) This is another chain rule, 1 f 0 (x) · f 0 (x) = f (x) ln(a) f (x) ln(a) 3 ! 7. Put each differential equation in the form: (f (x) + c)0 = k(f (x) + c) (a) f 0 (x) = 4 − 2f (x) f 0 (x) = −2(f (x) − 2) ⇒ (f (x) − 2)0 = −2(f (x) − 2) EXTRA: The general solution is f (x) = Ae−2x + 2 (b) f 0 (x) = 1 f (x) 10 −2 f 0 (x) = 1 (f (x) − 20) 10 (f (x) − 20)0 = 1 (f (x) − 20) 10 EXTRA: The general solution is f (x) = Ae1/10 x + 20 (c) f 0 (x) = 7 − 1 f (x) 30 f 0 (x) = − 1 (f (x) − 210) 30 (f (x) − 210)0 = − 1 (f (x) − 210) 30 1 EXTRA: The general solution is f (x) = Ae− 30 x + 210 8. Give the general solution for each differential equation (assume y is a function of t): NOTE: The general solution to y 0 = ky is y = Aekt , which you can verify. (a) y 0 = −2y (b) y 0 = −2y + 4 1 (c) y 0 = − 10 y y = Ae−2t (y − 2)0 = −2(y − 2) 1 y = Ae− 10 t 4 ⇒ y − 2 = Ae−2t ⇒ y = Ae−2t + 2