Solving Equations with e and ln x
1 10x2 1 1 y = 10x2 + 1 10x2 1 It’s a good idea to check our work by plugging y = 10x 2 +1 10x2 1 back into the original equation 3 2lny = ln(y + 1) + x Once again, we apply the inverse function ex to both sides We could use the identity e2lny = (elny)2 or we could handle the coe cient of 2 as shown below 2lny = ln(y + 1) + x lny2 = ln(y
Algebraic Properties of ln(
I = 1 2 ln(x 2 + 1) ln(x) Example 2 Express as a single logarithm: lnx + 3ln(x + 1) 1 2 ln(x + 1): I We can use our four rules in reverse to write this as a single
Limits involving ln(
x1 lnx = 1; lim x0 lnx = 1 : I We saw the last day that ln2 > 1=2 I Using the rules of logarithms, we see that ln2m = mln2 > m=2, for any integer m I Because lnx is an increasing function, we can make ln x as big as we choose, by choosing x large enough, and thus we have lim x1 lnx = 1: I Similarly ln 1 2n = nln2 < n=2 and as x approaches
AP Calculus BC - College Board
1 3 x < for x
Jiwen He 11 Geometric Series and Variations
Power Series Expansion of ln(1+x) Note: d dx ln(1+x) = 1 1+x = X∞ k=0 (−1)kxk for x < 1 Integration: ln(1+x) = X∞ k=0 (−1)k k +1 xk+1(+C = 0) = X∞ k=1 (−1)k k xk = x− 1 2 x2 + 1 3 x3 − 1 4 x4 +··· The interval of convergence is (−1,1] At x = 1, ln2 = X∞ k=1 (−1)k k = 1− 1 2 + 1 3 − 1 4 +··· Power Series
59 Representations of Functions as a Power Series
x n+1 n+ 1 = X 0 X ( 1) x +2 n+ 1 3 f(x) = ln(x2 + 1) This series is similiar to ln(x+ 1), except we replaced x with x2 So all we need to do is replace x with x2 in our power series representation for ln(x+ 1) from part (1) f(x) = ln(x2 + 1) = X 0 ( 1)n (x2) n+1 n+ 1 = X 0 ( 1)n x2 +2 n+ 1 4 f(x) = ln(1 x) Again, recall that we know ln(1
Techniques of Integration
cos2 x = 1−sin2 x sec2 x = 1+tan2 x tan2 x = sec2 x −1 If your function contains 1−x2, as in the example above, try x = sinu; if it contains 1+x2 try x = tanu; and if it contains x2 − 1, try x = secu Sometimes you will need to try something a bit different to handle constants other than one EXAMPLE10 2 2 Evaluate Z p 4− 9x2 dx We
ln(1+ x
x∈ R Example 1 Relative Rates of Growth and Decay Prove the following limit (1) lim x→∞ lnx x =0 Proof We may assume that x>1throughout the proof Now we claim that (2) 0
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