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How can I solve this equation generally? I can solve it by checking in my calculator. But I don't know any generalized way.

$$127=1+2+2^2+2^3+....+ 2^{x-1}$$

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Have you heard of geometric series? –  String May 8 at 12:18
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Consider the idea $(\alpha^0+\alpha^1+...+\alpha^{n-1})\cdot(\alpha-1)=\alpha^n-1$. It will work! –  String May 8 at 12:20
    
Yes ,I have heard –  Fazla Rabbi Mashrur May 8 at 12:21
    
If you succeed from the hints I gave, then post an answer to your own question ;) –  String May 8 at 12:22
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@String why not make it an answer? –  Ruslan May 8 at 14:09

4 Answers 4

up vote 2 down vote accepted

The most intuitive way: \begin{align*} &\;\underbrace{1 + 1} +2+2^2+2^3+ \cdots + 2^{x-1} \\ &= \underbrace{2 + 2} + 2^2 + 2^3 + \cdots + 2^{x-1} \\ &\;\;\;= \underbrace{4 + 2^2} + 2^3 + \cdots + 2^{x-1} \\ & \; \\ & \; \\ %vertical space &\quad \quad\quad \quad\quad \; \; \; \; \; \cdots \\ & \; \\ & \; \\ %vertical space &\quad \quad \quad \quad = \underbrace{2^{x-1} + 2^{x-1}} \\ &\quad \quad \quad \quad \quad \quad = 2^x \\ \end{align*} Hence, $$ 1 + 2 + 2^2 + 2^3 + \cdots + 2^{x-1} = 2^x - 1. $$

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Let $S=1+2+2^2+2^3+....+ 2^{x-1}$, then $2S=2+2^2+2^3+....+ 2^{x}$. So $2S-S=S=2^{x}-1$. Can you take it from here?

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Hint: $127=a\cdot \dfrac{r^{\text{number of terms}}-1}{r-1}$, where $a=$starting term, $r=$common ratio

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Prerequisites

Expanding $(\alpha^0+\alpha^1+...+\alpha^{n-1})\cdot(\alpha-1)$ in a clever way yields $$ \alpha\cdot(\alpha^0+\alpha^1+...+\alpha^{n-1})-1\cdot(\alpha^0+\alpha^1+...+\alpha^{n-1}) $$ where the trained eye sees that everything but $\alpha^n$ from the first product and $-1$ from the second cancels. Thus it follows (dividing by $\alpha-1$) that $$ (\alpha^0+\alpha^1+...+\alpha^{n-1})=\frac{\alpha^n-1}{\alpha-1} $$ which is probably found in your book under the subject of geometric series.

Solution

Applying the above with $\alpha=2$ shows that $$ f(x)=1+2+2^2+...+2^{x-1}=\frac{2^x-1}{2-1}=2^x-1 $$ so to solve $f(x)=127$ we have $2^x=128\iff x=\frac{\log(128)}{\log(2)}=7$.

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