Solve: $2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1\Bigl) = 0$ The question says to find the value of $x$ if, $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= 0$$
My approach:
I rewrote the expression as, 
 $$2^x\Bigl(2^x-1\Bigl) + \frac{2^x}{2}\Bigl(\frac{2^x}{2} -1 \Bigl) + .... + \frac{2^x}{2^{99}} \Bigl(\frac{2^x}{2^{99}} - 1 \Bigl)= 0$$
I then took $\bigl(2^x\bigl)$ common and wrote it as, 
$$2^x \Biggl[ \Bigl(2^x - 1\Bigl) + \frac{1}{2^1}\Bigl(2^x -2^1\Bigl) + \frac{1}{2^2}\Bigl(2^x - 2^2\Bigl) + \;\ldots + \frac{1}{2^{99}} \Bigl(2^x - 2^{99}\Bigl)\Biggl] = 0$$ 
After further simplification I got, 
$$\frac{2^x}{2^{99}} \Biggl[ \Bigl(2^x\cdot2^{99} - 2^{99}\Bigl) + \Bigl(2^x \cdot 2^{98} - 2^{99}\Bigl) + \ldots + \bigl(2^x -2^{99}\bigl)\Biggl] = 0$$ 
Taking $-2^{99}$ common I got, 
$$-2^x \Biggl[ \Bigl( 2^{x+99} + 2^{x+98} + \ldots + 2^{x+2} + 2^{x+1} + 2^x \Bigl)\Biggl]= 0$$
Now the inside can be expressed as $$\sum ^ {n= 99} _{n=1} a_n$$ Where $a_n$ are the terms of the GP.
Thus we can see that either $$-2^x= 0$$ Or, $$\sum ^ {n= 99} _{n=1} a_n = 0$$ 
Since the first condition is not poossible, thus, $$\sum ^ {n= 99} _{n=1} a_n = 0$$
So, 
$$2^{x + 99} \Biggl(\frac{1-\frac{1}{2^{100}}}{1-\frac{1}{2}} \Biggl) = 0$$ 
Either way once I solve this, I am not getting an answer that is even in the options. The answers are all in the form of logarithmic expressions.
Any help would be appreciated. We have to find the value of $x$.
 A: $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= $$ 
$$2^{2x}-2^x + 2^{2x-2}-2^{x-1} + .... + 2^{2x-198}-2^{x-99}= $$ 
$$(2^{2x}+ 2^{2x-2} .... + 2^{2x-198})-(2^x +2^{x-1} + .... + 2^{x-99})= $$ 
$$2^{2x-198}(2^{198}+ 2^{196} .... + 2^{0})-2^{x-99}(2^{99} +2^{98} + .... + 2^{0})= $$
$$2^{2x-198}{2^{200}-1\over 2^2-1} -2^{x-99}{2^{100}-1\over  2-1}=0 $$
If we cancel this with $2^{x-99}(2^{100}-1)$ we get
$$2^{x-99}{2^{100}+1\over 3} -1=0 $$
So $$2^{x-99} = {3\over 2^{100}+1}$$ and thus $$x= 99+\log_{2} {3\over 2^{100}+1}$$
A: Hint.
$$
2^{2x}\sum_{k=0}^{99}2^{-2k} =2^x\sum_{k=0}^{99}2^{-k}
$$
hence
$$
2^x = \frac{\sum_{k=0}^{99}2^{-k}}{\sum_{k=0}^{99}2^{-2k}} = 1.5\to x = 0.5849625007211562
$$
A: From scratch, the quantity you are looking at is $$\sum_{k=0}^{99} 2^{2(x-k)}-2^{x-k}=4^x\sum_{k=0}^{99}4^{-k}-2^x\sum_{k=0}^{99}2^{-k}=4^x\cdot\frac{4^{-100}-1}{-\frac34}-2^x\frac{2^{-100}-1}{-\frac12}=\\=2^x\left(2^x\cdot \frac{4-4^{-99}}{3}-2+2^{-99}\right)$$
That quantity is $0$ if and ony if $$2^x=\frac{3\cdot 2}4\cdot\frac{1-2^{-100}}{1-4^{-100}}=\frac{3}{2(1+2^{-100})}\\ x=\frac{\ln 3-\ln(1+2^{-100})}{\ln 2}-1$$
A: \begin{align}
2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl) &= 0 \\
\sum_{i=0}^{99}2^{x-i}\left(2^{x-i}-1\right) &= 0 \\
\sum_{i=0}^{99}\left[\left(2^{x-i}\right)^2-2^{x-i}\right] &= 0 \\
\sum_{i=0}^{99}\left(2^{x-i}\right)^2-\sum_{i=0}^{99}2^{x-i} &= 0 \\
\sum_{i=0}^{99}2^{2x-2i}&=\sum_{i=0}^{99}2^{x-i} \\
\dfrac{2^{2x}}{\sum_{i=0}^{99}2^{2i}}&=\dfrac{2^{x}}{\sum_{i=0}^{99}2^{i}} \\
2^x&=\dfrac{\sum_{i=0}^{99}2^{2i}}{\sum_{i=0}^{99}2^{i}} \\
&= \dfrac{4^{100}-1}{4-1}\cdot\dfrac{2-1}{2^{100}-1} \\
&= \dfrac{4^{100}-1}{3\left(2^{100}-1\right)}\\
x &= \log_2\left(\dfrac{4^{100}-1}{3\left(2^{100}-1\right)}\right)
\end{align}
A: Here, you forgot red: 
$$2^x \Biggl[ \Bigl(2^x - 1\Bigl) + \frac{1}{2^1\cdot \color{red}{2}}\Bigl(2^x -2^1\Bigl) + \frac{1}{2^2\color{red}{2^2}}\Bigl(2^x - 2^2\Bigl) + \;\ldots + \frac{1}{2^{99}\color{red}{2^{99}}} \Bigl(2^x - 2^{99}\Bigl)\Biggl] = 0$$ 
