# Math Help - logarithm

1. ## logarithm

could you possibly help me out? I am stuck on two questions
the general term of an infinite sereis is tn= ( pie square) (3/11)^ n+1. how many terms must be added to reach a sum that is greater than 0.32?
t1 = (pie square) (3/11)2 = 0.233672181
r = (3/11)
Sm = a * (1-rm)/(1-r)
0.32 * 8/11 = 0.233672181 * (1-rm)
I know I have to use logarithm but I get confused!

my second question!

convert 0.12 ( repeating decimal ) to a fraction using algebraic method!

I know that I would use the fomula :
s= a/ 1-r or am I totally wrong? can you guiys explain to me how I woudl do this question?

2. Hello, Andrew!

I know two ways to solve #2 . . .

Convert $0.\overline{12}$ to a fraction using algebraic methods.

Method #1: Geometric Series

We have: . $S\:=\:0.121212\hdots \:=\:0.12 + 0.0012 + 0.000012 + \hdots$

That is: . $S \:=\:\frac{12}{10^2} + \frac{12}{10^2} + \frac{12}{10^6} + \hdots$

We have a geometric series with first term $a = \frac{12}{100}$ and common ratio $r = \frac{1}{100}$

Its sum is: . $S\:=\:\frac{\frac{12}{100}}{1 - \frac{1}{100}}\:=\:\frac{\frac{12}{100}}{\frac{99} {100}} \:=\:\frac{12}{99} \:=\:\boxed{\frac{4}{33}}$

Method #2: .Let $x \:=\:0.12121212\hdots$

Multiply by 100: . $100x \:=\:12.12121212\hdots$
. . . Subtract $x:\qquad x \:=\:\;\;0.12121212\hdots$

. .And we have: . . $99x \:=\:12\quad\Rightarrow\quad x \:=\:\frac{12}{99} \:=\:\boxed{\frac{4}{33}}$

3. ## thank you....

thanks! it totally makes sense!

4. ## can you...

can you possiby hepl me with my first question? thanks!

could you possibly help me out? I am stuck on two questions
the general term of an infinite sereis is tn= ( pie square) (3/11)^ n+1. how many terms must be added to reach a sum that is greater than 0.32?
t1 = (pie square) (3/11)2 = 0.233672181
r = (3/11)
Sm = a * (1-rm)/(1-r)
0.32 * 8/11 = 0.233672181 * (1-rm)
I know I have to use logarithm but I get confused!

6. Hello, Andrew!

There must a typo in the first problem . . .

the general term of an infinite sereis is tn= ( pie square) (3/11)^ n+1.
how many terms must be added to reach a sum that is greater than 0.32?

It says: . $t_n \:=\:\pi^2\left(\frac{3}{11}\right)^{n+1}$

But the first term: $t_1\:=\:\pi^2\left(\frac{3}{11}\right)^2 \:=\:0.734...$
. . is already greater tha 0.32.

t1 = (pie square) (3/11)2 = 0.233672181

Here you used just $\pi$, not $\pi^2$.
So I'll assume that the "square" is a typo.
[By the way, it's not "pie" . . . it's "pi".]

We have: . $t_n \:=\:\pi\left(\frac{3}{11}\right)^{n+1}$ . . . and: . $t_1\:=\:\pi\left(\frac{3}{11}\right)^2\:=\:\frac{9 \pi}{121}$

Sum of a geometric series: . $S_m \:=\:a\cdot\frac{1 - r^m}{1 - r}$

So we have: . $S_m\:=\:\frac{9\pi}{121}\cdot\frac{1 - \left(\frac{3}{11}\right)^m}{1 - \frac{3}{11}} \:=\:\frac{9\pi}{88}\left[1 - \left(\frac{3}{11}\right)^m\right]$

We want this sum to exceed 0.32, so we have:
. . $\frac{9\pi}{88}\left[1 - \left(\frac{3}{11}\right)^m\right]\;>\;0.32$ . . . and we must solve for $m.$

Divide by $\frac{9\pi}{88}:\;\;1 - \left(\frac{3}{11}\right)^m \;> \:0.995956266$

Subtract 1: . $-\left(\frac{3}{11}\right)^m \;> \:-0.004043724$

Multiply by $-1:\;\;\left(\frac{3}{11}\right)^m \:<\:0.004043734$ . . .
reverse the inequality!

Take logs: . $\log\left(\frac{3}{11}\right)^m \:< \:\log(0.004043724)$

. . . . . . . $m\cdot\log\left(\frac{3}{11}\right) \;< \;\log(0.004043724)$

Hence:. . . . . . . . . . $m \;> \;\frac{\log(0.004043734)}{\log(3/11)} \;=\;4.241252175$
. .
Note: log(3/11) is negative, so we must reverse the inequality again.

Therefore: . $\boxed{m \,=\,5}$

For the sum to exceed 0.32, add the first five terms.