Logarithm and infinite geometric series  TOPIC_SOLVED

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.

Logarithm and infinite geometric series

Postby cruxxfay on Sat Nov 24, 2012 2:03 pm

"Consider the infinite geometric series

1 + (2x/3) + (2x/3)^2 + (2x/3) ^3 +...

For what values of x does the series converge?"

I know convergence means the sum tends to a finite series, so I have used the equation Sn = u/(1-r) where u = first term and r = ratio

Sn = 1/(1-2x/3)

But then I have no idea what to do next. I know (1- 2x/3) cannot be 0.

Second question is

"Solve 2(5^(x+1)) = 1 + 3/(5^x), giving the answer in the form a + log(base 5) b, where a and b are integers"

I managed to solve for x, but I have no idea how to put it in the form a + log(base 5) b

I got to 5^x = 3/5 or 5^x = -0.5
so x would just be the log (base 5) 3/5 (because it can't be negative)

Thanks for your time and effort
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Re: Logarithm and infinite geometric series

Postby theshadow on Sat Nov 24, 2012 6:14 pm

Did they give you a rule where in converges when |r|<1? Then find when |(2x)/3|<1.
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Re: Logarithm and infinite geometric series

Postby cruxxfay on Sun Nov 25, 2012 12:31 pm

No, that is the entire question.
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Re: Logarithm and infinite geometric series

Postby nona.m.nona on Mon Nov 26, 2012 3:26 am

cruxxfay wrote:No, that is the entire question.

I suspect that the previous reply is referring to your textbook and your classroom lecture notes. Were you given rules for working with these series?
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Re: Logarithm and infinite geometric series

Postby cruxxfay on Mon Nov 26, 2012 5:04 am

I realized I could use the ratio test to solve that sequence, although I have yet to learn it at school.

But then I am still confused with the a + log (base5) b part

In fact this is what I got:

5^x = 3/5
x = log(base5)^3/5
= log(base5)^3 - log(base5)^5
= -1 + log(base5)^3

so a = -1 and b = 3? I don't have the answer so I am not quite sure.
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Re: Logarithm and infinite geometric series  TOPIC_SOLVED

Postby theshadow on Mon Nov 26, 2012 1:55 pm

cruxxfay wrote:Second question is

"Solve 2(5^(x+1)) = 1 + 3/(5^x), giving the answer in the form a + log(base 5) b, where a and b are integers"

I managed to solve for x, but I have no idea how to put it in the form a + log(base 5) b.... In fact this is what I got:

5^x = 3/5
x = log(base5)^3/5
= log(base5)^3 - log(base5)^5
= -1 + log(base5)^3

so a = -1 and b = 3? I don't have the answer so I am not quite sure.

When you say log(base5)^5, do you mean log5(5)? I think so, and I get the same answer then.
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