Logarithm and infinite geometric series

[cruxxfay]

"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

[theshadow]

Did they give you a rule where in converges when |r|<1? Then find when |(2x)/3|<1.

[cruxxfay]

No, that is the entire question.

[nona.m.nona]

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?

[cruxxfay]

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.

[theshadow]

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.