Question

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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 1

1. No

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

Explanation:

Question 1

Given sequence:

`t(n)=5 \cdot 2^n`

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

`\implies 5 \cdot 2^n=200`

`\implies 2^n=40`

`\implies \ln 2^n=\ln 40`

`\implies n\ln 2=\ln 40`

`\implies n=(\ln 40)/(\ln 2)`

`\implies n=5.3219280...`

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:

`f(x)=5 \cdot 2^x`

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

`\implies 5 \cdot 2^x=200`

`\implies 2^x=40`

`\implies \ln 2^x=\ln 40`

`\implies x=(\ln 40)/(\ln 2)`

`\implies x=5.3219280...`

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

`x=(\ln 40)/(\ln 2)`

Answer 2

• 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)