Starting with the graph of y = e^x, find the equation of the graph that results

Question

Starting with the graph of y = e^x, find the equation of the graph that results from reflecting about the line y = 5. a. y = -e^x b. y = -e^x + 10 c. y = -e^x + 10 d. y = e^-x + 10 e. y = -e^-5x + 10 16. Solve each equation for x. a) In x = 2 b) e^e = 3 a. x = e^2, x = In(In 3) b. x = e^2, x = In(In 2) c. x = e^In 2/2, x = In(In 2) d. x = e^In2, x = e In 3 e. x = 2 In 2, x = In 3 17. Suppose the graph of f(x) = x^2 and f(x) = 2^x are drawn on a coordinate grid where the unit of measurement is 1 inch. At a distance 1 ft to the right of the origin, the height of the graph of f is 12 ft. Find the height of the graph of g. a. 331 ft b. 341 ft c. 171 ft d. 3,410 ft e. 3,110 ft 18. If f(d) = 3x + In x, find f^-1(3). a. f^-1 (3) = 1 b. f^-1 (3) = 2 c. f^-1 (3) = -1 d. f^-1 (3) = 0 e. f^-1 (3) = 3 19. If a bacteria population starts with 100 bacteria and doubles every three hours, then the number of bacteria after t hours is n = f(t) = 100(2^t/3). What will the population reach 50,000?

Explanation / Answer

15. c) y=-ex+10

As at x= ln5 y must be 5,as it is the point of reflection

16. a) x=ln2, x=ln(ln3)

by simple log rules.

17. At a distance of 1 ft means at x= 12 inches away,

f(x)=12ft=144

therefore,f(x)=x2 , so g(x)=2x,

therefore g(12)=212=4096 inches=341 ft.

thus, b)341ft.

18. clearly it can be seen that f(1)=3 ,

therefore a) f-1(3)=1.

19. when population becomes 50000,

we use,

50000=100x2t/3

500=2t/3

log500=(t/3)log2

t=26.897h.

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