04)ln(ln(x)) = 1ln(x) = e^1ln(x) = ex = e^ee^(ax) = C * e^(bx)e^(ax) / e^(bx) = Ce^(ax - bx) = Ce^(x * (a - b)) = Cx * (a - b) = ln(C)x = ln(C) / (a - b)5)arcsin(sqrt(3)/2) =>arcsin(sin(pi/3)) =>pi/3arccos(-1) =>arccos(cos(pi)) =>pi6)arctan(1 / sqrt(3)) =>arctan(tan(pi/6)) =>pi/6arcsec(2) =>arcsec(sec(pi/4)) =>pi/47)arctan(1) =>arctan(tan(pi/4)) =>pi/4arcsin(1 / sqrt(2)) =>arcsin(sin(pi/4)) =>pi/48)arccot(1 / sqrt(3)) =>arccot(cot(pi/3)) =>pi/3arccos(-1/2) =>arccos(cos(-pi/3)) =>-pi/39)tan(arctan(10)) =>10arcsin(sin(7pi/3)) =>arcsin(sin(pi/3)) =>pi/310)tan(arcsec(4))sec(arcsec(4))^2 - tan(arcsec(4))^2 = 14^2 - tan(arcsec(4))^2 = 115 = tan(arcsec(4))^2+/- sqrt(15) = tan(arcsec(4))sin(2 * arcsin(3/5)) =>2 * sin(arcsin(3/5)) * cos(arcsin(3/5)) =>2 * (3/5) * cos(arcsin(3/5)) =>(6/5) * cos(arcsin(3/5))cos(arcsin(3/5))^2 + sin(arcsin(3/5))^2 = 1cos(arcsin(3/5))^2 + (3/5)^2 = 1cos(arcsin(3/5))^2 = 16/25cos(arcsin(3/5)) = +/- 4/5(6/5) * (+/- 4/5) =>+/- 24 / 25-1) 2 feet to the right of the origin is 24 inches. Therefore, find f(24) and g(24).2) a)Put the power inside the natural log:= e^(ln(25))Then cancel it with the e (inverse functions):= 25b) Cant quite tell what the powers are on, but it probably cancels to just 10.3) Let y = e^x,......
10 * ln(e) =>
10 * 1 =>
10
3)
e^(2x) - 3 * e^(x) + 2 = 0
e^(x) = t
t^2 - 3t + 2 = 0
t = (3 +/- sqrt(9 - 4 * 1 * 2)) / (2 * 1)
t = (3 +/- sqrt(9 - 8)) / 2
t = (3 +/- sqrt(1)) / 2
t = (3 +/- 1) / 2
t = 4/2 , 2/2
t = 2 , 1
e^(x) = 2 , 1
x = ln(2) , ln(1)
x = ln(2) , 0
4)
ln(ln(x)) = 1
ln(x) = e^1
ln(x) = e
x = e^e
e^(ax) = C * e^(bx)
e^(ax) / e^(bx) = C
e^(ax - bx) = C
e^(x * (a - b)) = C
x * (a - b) = ln(C)
x = ln(C) / (a - b)
5)
arcsin(sqrt(3)/2) =>
arcsin(sin(pi/3)) =>
pi/3
arccos(-1) =>
arccos(cos(pi)) =>
pi
6)
arctan(1 / sqrt(3)) =>
arctan(tan(pi/6)) =>
pi/6
arcsec(2) =>
arcsec(sec(pi/4)) =>
pi/4
7)
arctan(1) =>
arctan(tan(pi/4)) =>
pi/4
arcsin(1 / sqrt(2)) =>
arcsin(sin(pi/4)) =>
pi/4
8)
arccot(1 / sqrt(3)) =>
arccot(cot(pi/3)) =>
pi/3
arccos(-1/2) =>
arccos(cos(-pi/3)) =>
-pi/3
9)
tan(arctan(10)) =>
10
arcsin(sin(7pi/3)) =>
arcsin(sin(pi/3)) =>
pi/3
10)
tan(arcsec(4))
sec(arcsec(4))^2 - tan(arcsec(4))^2 = 1
4^2 - tan(arcsec(4))^2 = 1
15 = tan(arcsec(4))^2
+/- sqrt(15) = tan(arcsec(4))
sin(2 * arcsin(3/5)) =>
2 * sin(arcsin(3/5)) * cos(arcsin(3/5)) =>
2 * (3/5) * cos(arcsin(3/5)) =>
(6/5) * cos(arcsin(3/5))
cos(arcsin(3/5))^2 + sin(arcsin(3/5))^2 = 1
cos(arcsin(3/5))^2 + (3/5)^2 = 1
cos(arcsin(3/5))^2 = 16/25
cos(arcsin(3/5)) = +/- 4/5
(6/5) * (+/- 4/5) =>
+/- 24 / 25
1) 2 feet to the right of the origin is 24 inches. Therefore, find f(24) and g(24).
2) a)Put the power inside the natural log:
= e^(ln(25))
Then cancel it with the e (inverse functions):
= 25
b) Can't quite tell what the powers are on, but it probably cancels to just 10.
3) Let y = e^x, you'll end up with a quadratic [e^(2x) = (e^x)²]. Solve that, then put e^x back and take the natural log of both sides.
4) a) make both sides powers of e's to cancel the natural log. Do it again to find your answer.
b) Divide by e^(ax), so you can just subtract the powers. Factor out the a, do some algebra, take the natural log, and do some more algebra to solve for x.
5 - 10 These are all special angles, you should be able to compute them if you know the special angles of the trig functions, or you can just punch them into any calculator.
