Re: Geometric Sequences 2

If t1 = 100, that means after zero washings. Some people might write "t0" for that, but that's a matter of convention. If t1 = 100, then t2 = 95 (after one washing), and t11 = 59.87, after 10 washings. You need to plug 11 in for n, not ten. Just a matter of how the counting is done.

Part e requires logarithms. If you're doing this, I suppose you're familiar with logarithms. Go back and review them. If you have problems, ask further questions.

Re: Geometric Sequences 2

I'm in high school grade 11, and I have actually never been taught to do logs, I'm just doing them now...

I understand d) and I've tried doing e) again:

tn = (100)(0.95^n-1)

25 = (95^n-1)

(log(25))=n-1(log(95))

1.39794001 = n-1 (1.977723601)

1.39794001/(1.977723601)= n-1

1.39794001/(1.977723601)+1 = n

n = 1.71

The answer is after 27 washes...

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Re: Geometric Sequences 2

Your answer to 3) is correct, I think, but I don't follow your figuring. Did you maybe leave something out? For instance, I got

n = 28.027

which corresponds to 27 washings. You do need to subtract one from the value of n to get the "washing number." That's a consequence of counting the way you do. If you started counting from 0, it might be easier.

You might want to glance at the attached pdf.

You probably don't know what natural logs are yet (notation: ln). Learning about that would probably be an enrichment activity for you, not necessary for your class.

Re: Geometric Sequences 2

Quote:

Originally Posted by

**Mathnood768** I'm in high school grade 11, and I have actually never been taught to do logs, I'm just doing them now...

I understand d) and I've tried doing e) again:

tn = (100)(0.95^n-1)

25 = (95^n-1)

This is incorrect. You should have 0.25= (0.95)^{n-1}

Quote:

(log(25))=n-1(log(95))

1.39794001 = n-1 (1.977723601)

1.39794001/(1.977723601)= n-1

1.39794001/(1.977723601)+1 = n

n = 1.71

The answer is after 27 washes...

I presume this is the answer "in the back of the book".