need help with this notation problem

[Blucas22]

well I found out the answer but I need to know if there is another way to finding out the the "k" or "6" without using logs if its possible

Question: Express the series in sigma notation: 4/3 + 2 + 3.... 81/8

this is the answer to the sigma notation


6
(sigma) (4/3)(3/2)^n-1
n=1


here is my work to finding "6" in this problem: lines 1-7

line 1: 81/8=(4/3)*(3/2)^k-1 (divide both sides by (4/3) )

line 2: 243/32=(3/2)^k-1 (log both sides)

line 3: log(243/32)=log(3/2)^k-1 (bring down the k-1)

line 4: log(243/32)=(k-1)*log(3/2) (divide both sides by log(3/2)

line 5: [log(243/32)]/[log(3/2)]=k-1 (add 1 to both sides)

line 6: [log(243/32)]/[log(3/2)] + 1 = k

line 7: 5 + 1 = k k=6


here is where i need help... I'm in college and i would of approached this with logs like i did.. i found the answer but I am helping my sister out with her hw and I'm not sure if they used logs yet. I looked at problems in her book and they use same bases all the time.. if you look above at line 2 i tried not to use logs... I took out 243 from the left side and 3 from the right. so the problem would look like this (243)*(1/32)=(3)*(1/2)^k-1.next i divided both sides by 3 and got this;
81*(1/32)=(1/2)^k-1. now i have same bases of 2 on both sides. this is where i need help. now i set up the problem like this--- 81*(2^-5)=(2^-1)^k-1 is it possible to solve this using same bases or any other way without using logs.. if you could please help i would appreciate it.


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[morson]

No need to take logs.

That series looks to be geometric with common ratio 1.5.

\(\displaystyle \sum_{n=1}^{j} [4/3 (3/2)^{n-1}]\\)

Now it's a matter of finding the value of j. We know that the jth term of this series is:

\(\displaystyle t_j = \frac{4}{3}\ (\frac{3}{2}\)^{j-1}\)

Putting \(\displaystyle t_j = \frac{81}{8}\\), we have:

\(\displaystyle \frac{81}{8}\ = \frac{4}{3}\ (\frac{3}{2}\)^{j-1}\), hence \(\displaystyle (\frac{3}{2}\)^{j-1} = \frac{243}{32}\\)

Hence \(\displaystyle (\frac{3}{2}\)^{j-1} = (\frac{3}{2}\)^{5}\)

So, \(\displaystyle j - 1 = 5\), hence \(\displaystyle j = 6\)

Therefore, the summation is:

\(\displaystyle \sum_{n=1}^{6} [4/3 (3/2)^{n-1}]\\), which is 665/24?

 

[pka]

\(\displaystyle \L\begin{array}{rcl}
\sum\limits_{n = 1}^K {\frac{4}{3}\left( {\frac{3}{2}} \right)^{n - 1} \Rightarrow \frac{4}{3}\left( {\frac{3}{2}} \right)^{K - 1} = \frac{{81}}{8}} \\
\Rightarrow \frac{{2^2 }}{3}\left( {\frac{3}{2}} \right)^{K - 1} & = & \frac{{3^4 }}{{2^3 }} \\
\Rightarrow \left( {\frac{3}{2}} \right)^{K - 1} & = & \frac{{3^5 }}{{2^5 }} \\
\Rightarrow K - 1 & = & 5 \\
\Rightarrow K & = & 6 \\
\end{array}\)

 

[Blucas22]

thanks everyone