Solution to an infinite series Given an infinite series in the form of:
$$a \cdot a^{2\log(x)} \cdot a^{4\log^2(x)} \cdot a^{8\log^3(x)} \dotsb  = \frac{1}{a^7} $$
find the solution for all positive and real $a$ other than $1$.
The textbook, from which this problem was taken, says that there are no solutions to this equation.
Given that I didn't know that before I attempted to solve it, I managed to scramble a solution which gives $x = 10^{4/7}$ and here's how I did it:
First, let's take the $\log_a$ from both sides and expand by the log multiplication rule. We then obtain:
$$
 1+2\log(x)+4\log^2(x)+8\log^3(x)+\dotsb = -7.  
$$
Let's now call the left hand side $S(x)$, so we have $S(x) = -7$. If we then go back to our last equation and subtract one from both sides and multiply by $\frac{1}{2\log(x)}$ we get:
$$  1+2\log(x)+4\log^2(x)+8\log^3(x)+\dotsb = \frac{-4}{\log(x)}.   $$
So we have $S(x)$ again on the left hand side, therefore we can set the two expressions equal to one another. After a quick algebraic rearrangement, we get:
$$ \log(x) = \frac{4}{7} \implies x = 10^{\frac{4}{7}} $$
I tend to believe the textbook rather than myself but don't know which of the steps was invalid. The method of substitution by $S(x)$ resembles a technique Euler supposedly used to solve some infinite fraction problems so it perhaps doesn't apply in this case with functions.
 A: Your approach to find $x$ was well and good. However before assuming we have found the ultimate solution to any problem, it is good practice to substitute your solution back in the original question and see if everything checks out.
That being said, notice that:
$$a^{2\log x}=a^{2\times\frac47}=a^{\frac87}$$
$$a^{4\log^2x}=a^{4\times\frac{16}{49}}=a^{64\over 49}$$
Focus on the  powers since we have anyway tackled the problem with $a$ by taking the logarithm. Now you can see that the powers go on increasing and each power is greater than $1$. This creates a divergent geometric series and hence can never converge to the "supposed" value of $-7$
This means that we were wrong in considering the solution to be $x=10^{\frac47}$. Since this is the only solution we could come up with, there should be no solutions for the value of $x$. And since there is no $x$ that satisfies the original equation, we cannot conclude on a value of $a$.
Hence, no $a$ under the given constrains, can satisfy the given equation. $\square$
A: Consider the exponent in the LHS of the original equation.
It is a geometric series with first term 1, common ratio $2 \times \frac47$.
The absolute value of the common ratio is greater than one, so the series does not have a finite sum. Thus, the solution you found is fallacious.
