[4 marks
c) log2X, log2 (X + 9), and log2 (x + 45) are three consecutive terms of an arithmetic
progression
i)
ii)
Find X
Find the 5th term as a single logarithm.
lanice​

[4 Marksc) Log2X, Log2 (X + 9), And Log2 (x + 45) Are Three Consecutive Terms Of An Arithmeticprogressioni)ii)Find

Answers

Answer 1

Answer:

1. x = 3

2. The 5th term = Log₂768.

Step-by-step explanation:

From the question given above, the following data were obtained:

Log₂x, Log₂(x + 9, Log₂(x + 45)

x =?

5th term (T₅) =.?

1. Determination of the value of x.

We shall determine the value of x as follow:

First term = Log₂x

2nd term = Log₂(x + 9)

3rd term = Log₂(x + 45)

Common difference = 2nd term – first term = 3rd term – 2nd term

Log₂(x + 9) – Log₂x = Log₂(x + 45) – Log₂(x + 9)

Recall

Log M – Log N = Log (M/N)

Therefore,

Log₂(x + 9) – Log₂x = Log₂(x + 9)/x

Log₂(x + 45) – Log₂(x + 9) = Log₂(x + 45) /(x + 9)

Thus:

Log₂(x + 9) – Log₂x = Log₂(x + 45) – Log₂(x + 9)

Log₂(x + 9)/x = Log₂(x + 45)/(x + 9)

Cancel Log₂ from both side

(x + 9)/x = (x + 45)/(x + 9)

Cross multiply

(x + 9)(x + 9) = x(x + 45)

x² + 9x + 9x + 81 = x² + 45x

x² + 18x + 81 = x² + 45x

Rearrange

x² – x² + 81 = 45x – 18x

81 = 27x

Divide both side by 27

x = 81/27

x = 3

Therefore, the value of x is 3.

2. Determination of the 5th term.

We'll begin by calculating the common difference (d)

x = 3

First term = Log₂x

First term = Log₂3

2nd term = Log₂(x + 9)

2nd term = Log₂(3 + 9)

2nd term = Log₂12

Common difference (d) = 2nd term – first term

Common difference (d) = Log₂12 – Log₂3

= Log₂(12/3)

Common difference (d) = Log₂4

Finally, we shall determine the 5th term as follow:

First term (a) = Log₂3

Common difference (d) = Log₂4

5th term (T₅) =.?

T₅ = a + 4d

T₅ = Log₂3 + 4Log₂4

Recall:

nLogM = LogMⁿ

Therefore,

4Log₂4 = Log₂4⁴ = Log₂256

T₅ = Log₂3 + 4Log₂4

T₅ = Log₂3 + Log₂256

Recall:

Log M + Log N = Log (M×N)

Thus,

T₅ = Log₂3 + Log₂256

T₅ = Log₂(3 × 256)

T₅ = Log₂768

Therefore, the 5th term is Log₂768


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[tex](3x +1)\cdot 2x - x^2 \cdot 3\\\\=2x\left(3x+1\right)-3x^2\\\\\mathrm{Expand}\:2x\left(3x+1\right):\quad 6x^2+2x\\\\=6x^2+2x-x^2\cdot \:3\\\\\mathrm{Simplify}\:6x^2+2x-x^2\cdot \:3:\\\quad 3x^2+2x[/tex]

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Answers

Answer:

The value is      [tex]y  =  \$ 225[/tex]

Step-by-step explanation:

From the question we are told that

   The amount charge per year is  [tex]k  =  \$ 200[/tex]

   The  number of members it will have at this amount is  [tex]n  =  50[/tex]

   The amount amount increase that will lead to the loss of a single member is   [tex]z =  \$ 5[/tex]

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      [tex]I  =  Amount \  due\ paid *  Number \  of members[/tex]

Now let x denote the number of member lost

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      [tex]I  =  (k + zx) (n-x )[/tex]

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=>         [tex] x   =  5[/tex]

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          [tex]I  =  (200 + 5 (5)) (50-5 )[/tex]

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i think it helps you

☽------------❀-------------☾

Hi there!

~

[tex]35-(9+8)\times 2=[/tex]

[tex]= 35 - (17)(2)[/tex]

[tex]= 35 - 34[/tex]

[tex]= 1[/tex]

❀Hope this helped you!❀

☽------------❀-------------☾

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Answers

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