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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.

Xander Henderson
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SierraTango
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2 Answers2

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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. $\square$

sai-kartik
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  • I saw your comment about the long-time user with a mickey avatar who posts PSQs. You know why? Because the same small group of users who protect PSQs have been protecting a lot of this user's PSQs by undeleting/reopening them. – user21820 Jul 23 '20 at 18:19
  • @user21820 so you're saying there's a whole group out there trynna live off from PSQ's? Something needs to be done about it. – sai-kartik Jul 24 '20 at 01:08
  • @user21820 also if you don't mind, could we move our discussion to chat? – sai-kartik Jul 24 '20 at 01:16
  • It's a complicated situation. A few of them indeed protect PSQs that they have answered. But some of them protect PSQs for a more reasonable stated reason that they just want to protect mathematical content without caring about whether the asker was lazy or not. Sure let's move to chat; you can find me in SBA's realm where we can discuss anything under the sun (or beyond). =) – user21820 Jul 24 '20 at 05:41
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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.

Benjamin Wang
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    Stated another way... Sierra showed if there is a solution, then it must be $10^{4/7}$. The other part (which fails in this case, as it does in some of Euler's phony results) is to show that there is a solution at all. – GEdgar Jun 18 '20 at 18:31