..READ THIS FIRST..Problems
about Exponential and Logarithmic functions
READ THIS FIRST
If a problem is solved. It is not 'the' answer.No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.
Problems about Exponential and Logarithmic functions
Level 1 problems
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Solve 15.3x+1 - 243.5x-2 = 0
3.5. 3x+1= 35. 5x-2 <=> 3x-3 = 5x-3 <=> x - 3 = 0 <=> x = 3
Level 2 problems
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Solve log4/x(x2 - 6) = 2
Since 4/x is the base of a log function x is > 0. The equation is then equivalent with (4/x)2 = x2- 6 <=> 16 = x4- 6x2 let x2 = t <=> 16 = t2- 6t <=> t=-2 or t=8 <=> x = sqrt(8) (since x >0) -
Calculate the first and second derivative of y = x3. e-xy' = 3x2. e-x- x3.e-x y" = 6x.e-x - 3x2. e-x - 3x2. e-x + x3.e-x y" = 6x.e-x - 6x2. e-x + x3.e-x -
Calculate the derivative of y = ln(tan(x/2))1 1 1 y' = --------- .--------------------.--- tan(x/2) cos(x/2).cos(x/2) 2 1 = ----------------------- 2 sin(x/2).cos(x/2) 1 = ------ sin(x) -
Calculate the derivative of y = ln(tan(x/2 + pi/4))Analogous as in the previous exercise, we find y' = 1/cos(x) -
The function f(x) is given by (ex-1)/x for all x not 0 1 for x = 0 Investigate if f(x) is continuous for x = 0f(x) is continuous for x = 0 <=> lim f(x) = f(0) 0 <=> lim f(x) = 1 0 ex - 1 Now, lim f(x) = lim ------- = 0 0 x ex = lim ---- = 1 0 1 Hence f(x) is continuous for x = 0 -
Find sin(x) + cos(x) - ex lim ---------------------- 0 ln(1+ x2)With l'Hospitals rule cos(x) -sin(x) - ex = lim -------------------- 0 2x/(1 + x2) cos(x) -sin(x) - ex = lim (1 + x2) . -------------------- 0 2x cos(x) -sin(x) - ex = lim (1 + x2) . lim -------------------- 0 0 2x - sin(x) - cos(x) - ex 1. lim -------------------------- = -1 0 2 -
Given : f(x) = ln(e-2 + ex)
Prove that f(x) increases for all x.
What is the equation of the inverse function?
f'(x) = (e-2 + ex)-1 . ex > 0 for all x.So, f(x) increases for all x.
The equation of the function f(x) is
y = ln(e-2 + ex) <=> ey = e-2 + ex <=> ex = e-2 - ey <=> x = ln(e-2 - ey)
The equation of the inverse function isy = ln(e-2 - ex)
Level 3 problems
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Solve x(2ln(x)-1) + e(1/9) = (1 + e(1/9)) x(ln(x)-0.5)Let y = x(ln(x)-0.5) Then the given equation becomes y2 + e(1/9) = (1 + e(1/9)) y This is a quadratic equation in y. The roots are y = 1 and y = e(1/9) a) y = 1 <=> x(ln(x)-0.5) = 1 <=> (ln(x) - 0.5) = 0 <=> ln(x) = 0.5 <=> x = sqrt(e) b) y = e(1/9) <=> x(ln(x)-0.5) = e(1/9) Taking ln of both sides, we have <=> (ln(x) - 0.5).ln(x) = 1/9 <=> ln(x).ln(x) - 0.5 ln(x) - 1/9 = 0 This is a quadratic equation in ln(x). The roots are ln(x) = 2/3 and ln(x) = -1/6 <=> x = e2/3 and x = e-1/6 -
Calculate lim x-x 0Instead of calculating the limit directly, we calculate ln lim x-x = lim ln(x-x) 0 -ln(x) = lim ( -x.ln(x)) = lim ------ 0 0 1/x With l'Hospitals rule we have 1/x = lim ------- = lim x = 0 1/(x2) So, -x -x ln lim x = 0 => lim x = 1 0 0 -
Calculate the derivative of xx
d We know that --(ln ( xx ) ) dx 1 d = ----. --( xx) xx dx d So, --( xx) dx x d = x . --(ln ( xx ) dx x d = x .--(x.ln ( x ) ) dx x = x .(ln(x) + 1) You can generalize this method to calculate the derivative of g(x) f(x) -
Find ln(ln(1+x4)) lim --------------- 0 ln(ln(1+x2))With l'Hospitals rule ln(1+x2) 1+x2 4x3 = lim ----------.------.------ 0 ln(1+x4) 1+x4 2x 2 2 2 x ln(1+x ) 1+x = 2 lim ----------.lim----- 0 4 0 4 ln(1+x ) 1+x x2.ln(1+x2) = 2 lim ------------- 0 ln(1+x4) x4.ln(1+x2) = 2 lim -------------- 0 x2.ln(1+x4 ) by properties of logarithm ln(1+x2)(1/x2) = 2 lim -------------------- 0 ln(1+x4)(1/x4) by properties of the number e ln e = 2 ------ = 2 ln e -
Investigate the function ex + 3 e-x y = ln(-----------------) ex + 1The argument of ln is strictly positive. The domain of the function is R. ex + 3 e-x y = 0 <=> ------------- = 1 ex + 1 <=> ex + 3 e-x = ex + 1 <=> 3 = ex <=> x = ln 3 ex - 6 - 3 e-x y' = ... = ---------------------------- (ex + 3 e-x) (ex + 1) e2 x - 6 ex - 3 = -------------------------- (e2 x + 3) (ex + 1) let ex= u then y' = 0 <=> u2- 6 u - 3 = 0 ___ <=> u = 3 + 2 V 3 __ <=> x = ln(3 + 2 V 3) Investigation of the sign of y' tells that there is a minimum for that x-value. Since the domain is R, there are no vertical asymptotes To find the horizontal asymptotes we calculate ex + 3 e-x lim ln(---------------) + infty ex + 1 x -x e 3 e = ln(lim -------- + lim -------- ) = ln (1 + 0 ) = 0 x x e + 1 e + 1 y = 0 is a horizontal asymptote for x -> + infty ex + 3 e-x lim ln(---------------) - infty ex + 1 x -x e 3 e = ln(lim -------- + lim -------- ) = ln (1 + infty) = + infty x x e + 1 e + 1 There is not a horizontal asymptote for x -> - infty To find the oblique asymptotes y = ax + b we calculate ex + 3 e-x a = lim ln(---------------) / x -infty ex + 1 With l'Hospitals rule we find e2 x - 6 ex - 3 = lim -------------------------- = -1 -infty (e2 x + 3) (ex + 1) b = lim f(x) + x -infty This limit is not so easy to calculate because of the presence of ln. Therefore we calculate b (f(x)+x) e = lim e -infty ex + 3 e-x = lim (--------------) ex ex + 1 e2x+ 3 = lim (-----------) = 3 ex + 1 It follows that b = ln 3. The oblique asymptote is y = -x + ln 3 -
Solve next system for all real solutions ex + e-y2 = 1 (1) e2x + sqrt( e- y2) = 1 (2)Let X = ex and Y = sqrt( e- y2) Then X > 0 and Y > 0 and 2.ln(Y) = - y2 < 0 Thus, 0 < Y < 1 With this substitution the system becomes / | X + Y2 = 1 (3) | | X2 + Y = 1 (4) \ (3) - (4) gives X - X2 + Y2 - Y = 0 <=> (X - Y) - (X2 - Y2) = 0 <=> (X - Y) ( 1 - X - Y ) = 0 <=> X = Y or Y = 1 - X First take Y = 1 - X . (4) becomes X2 + 1 - X = 1 This gives X = 0 or ( X = 1 and Y = 0 ) Both cases give no solution for the original system. Now take X = Y (3) becomes X2 + X - 1 = 0 The positive root is X = (sqrt(5) - 1)/2 So, X = Y = (sqrt(5) - 1)/2 x = ln ( (sqrt(5) - 1)/2 ) = -0.481 and e- y2 = Y2 = (3 - sqrt(5))/2 <=> - y2 = ln ( (3 - sqrt(5))/2 ) <=> y2 = ln ( 2/(3 - sqrt(5)) ) <=> y = sqrt ( ln ( 2/(3 - sqrt(5)) ) ) = 0.981 or y = - sqrt ( ln ( 2/(3 - sqrt(5)) ) ) = - 0.9810
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