meredith48034
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1. Is it possible for the sequence t(n) = 5·2ⁿ to have a term with the value of 200? If so, which term is it? If not, justify why not.

2. Is it possible for the function f(x) = 5·2ˣ to have an output of 200? If so, what input gives this output? If not, justify why not.



Answer :

mhanifa

Answer:

  • 1) No,
  • 2) Yes, x ≈ 5.32

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Part 1

Given sequence:

  • t(n) = 5 · 2ⁿ

If t(n) = 200, we can try to find the value of n:

  • 5 · 2ⁿ = 200
  • 2ⁿ = 40

There is no integer solution, since 32 < 40 < 64 or 2⁵ < 40 < 2⁶, the value of n should be between 5 and 6.

The sequence should include integer numbers, so there is no solution.

Part 2

Given function:

  • f(x) = 5 · 2ˣ

Solve for x if f(x) is 200:

  • 5 · 2ˣ = 200
  • 2ˣ = 40
  • log 2ˣ = log 40
  • x log 2 = log 40
  • x = log 40 / log 2
  • x = 5.32 (rounded)
semsee45

Answer:

1.  No

[tex]\textsf{2.} \quad x=\dfrac{\ln 40}{\ln 2} \approx5.32\;(\sf 2\;d.p.)[/tex]

Step-by-step explanation:

Question 1

Given sequence:

[tex]t(n)=5 \cdot 2^n[/tex]

To determine if the sequence has a term with a value of 200, substitute t(n)=200 into the equation and solve for n:

[tex]\implies 5 \cdot 2^n=200[/tex]

[tex]\implies 2^n=40[/tex]

[tex]\implies \ln 2^n=\ln 40[/tex]

[tex]\implies n\ln 2=\ln 40[/tex]

[tex]\implies n=\dfrac{\ln 40}{\ln 2}[/tex]

[tex]\implies n=5.3219280...[/tex]

In a sequence, n is a positive integer.  Therefore, it is not possible for the sequence to have a term with the value of 200, as when t(n)=200, n is not a positive integer.

Question 2

Given function:

[tex]f(x)=5 \cdot 2^x[/tex]

To determine if the function has an output of 200, substitute f(x)=200 into the function and solve for x:

[tex]\implies 5 \cdot 2^x=200[/tex]

[tex]\implies 2^x=40[/tex]

[tex]\implies \ln 2^x=\ln 40[/tex]

[tex]\implies x=\dfrac{\ln 40}{\ln 2}[/tex]

[tex]\implies x=5.3219280...[/tex]

Therefore, it is possible for the function to have an output of 200 when:

[tex]x=\dfrac{\ln 40}{\ln 2}[/tex]

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