Wrong log reasoning?

Question

From this question, we are given a system of equations and a restriction:

Let $a,b,c > 0$ for

$$\begin{cases} \log_a(b^x) & = & 2 \\ \log_b(c^x) & = & 2 \\ \log_c(a^x) & = & 5 \end{cases}$$

$x =\,\,?$

$$\log_a(b^x) = 2 \iff b^x=a^2$$

We now have $$ b^x = a^2 \\ c^x = b^2 \\ a^x = c^5 $$

Adding these up and rearranging, we have

$$a^x + b^x + c^x = a^2 + b^2 + c^5$$

Now $x$ does not seem to have a consistent solution. How is this possible?

A comment suggested this might be due to an over-determined system, but this answer shows that it has the unique solution $x^3 = 20$.

From the original post, we have the following question (which was not answered)

Looks to me like x is equal 2 different numbers at the same time which is strange, what am I doing wrong here?

Answer 1

The problem you're hitting is, as someone in a comment pointed out, you're assuming that: $a^x + b^y + c^z = a^m + b^n + c^o \implies x=m, y=n, z=o$, but this is not true. here's a concrete counterexample:

$$ 1^4 + 2^4 + 4^4 = 1^4 + 2^8 + 4^2$$

Both sides clearly evaluate to 273, use the same constants for $a,b,c$, yet the exponents of each constant are different.

Going back to the original equation in question, it is true that there is some $x$ - specifically $20^{1/3}$ - such that $a^x + b^x + c^x = a^2 + b^2 + c^5$. The error in your logic is that you're trying to partition the sum into 3 parts and say: "$a^x = a^2$, $b^x=b^2$, $c^x = c^5$" when it is not possible to claim that these partitions of the sums are equivalent.

In actuality, there really are three equal partitions in this equation already! They are in fact the first three equations you provided: $$b^x = a^2$$ $$c^x = b^2$$ $$a^x = c^5$$

Answer 2

You also need $a,b,c\ne1$ or the logarithms wouldn't make sense. Rather than summing up the equations after converting to exponentials, use $$ \log_pq=\frac{\log q}{\log p} $$ where the unadorned $\log$ means any base you prefer. Thus you have $$ \begin{cases} x\log b=2\log a \\[4px] x\log c=2\log b \\[4px] x\log a=5\log c \end{cases} $$ If you multiply the three equations, you get $$ x^3\log b\log c\log a=20\log a\log b\log c $$ and, since none of $a$, $b$ or $c$ is $1$, you can simplify.

The system is indeed underdetermined, because you can't solve it for $a$, $b$ and $c$. Namely, using $A=\log a$, $B=\log c$ and $C=\log c$, you have \begin{cases} 2A-xB=0 \\[4px] 2B-xC=0 \\[4px] xA-5C=0 \end{cases} which easily solves to $$ A=-\frac{x^2}{4}C,\qquad B=\frac{x}{2}{C} $$ and $C$ is a free variable (here $x=\sqrt[3]{20}$ as shown before).