Can the " same base, same exponent " rule be extended to sums?

Question

By " same base, same exponent " rule I mean :

$$b^x = b^n \iff x=n$$

Can the equation: $$ 2^x+4^x = 6$$ be solved by saying $$ 2^x+4^x = 6 = 2+4= 2^1 + 4^1$$ and consequently, that $$x = 1? $$

Answer 1

If one of the bases is less than one, you can get, for example $$0.5^x+2^x =2.5$$ which has two solutions $x=1$ and $x=-1$

Answer 2

No.

It's better: $$(2^x)^2+2^x-6=0$$ or $$(2^x-2)(2^x+3)=0$$ or $$2^x=2,$$ which gives $$x=1.$$ Also, you can say that $2^{2x}+2^x$ increases, which by your work gives $x=1$ again.

Answer 3

Well, this is one way to obtain a solution, especially if you don't know any other way to proceed -- we usually say the solution has been obtained by observation.

However, the drawback with this method is that it may not work with a sufficiently complicated equation. Also, it doesn't guarantee that one has found all the solutions -- although one may ascertain this by other means.

In this case you can solve in a more sure way, as one of the answers has demonstrated, by noticing that this is quadratic in $2^x.$

But once again, observation is one way to proceed when nothing else works.

Answer 4

A perhaps more worrisome example is that $$ (-1)^a = (-1)^b $$ has infinitely many solutions in integers. The equation holds if $a$ and $b$ are both even and it holds if $a$ and $b$ are both odd. This is very far from requiring $a = b$.

Answer 5

You should be careful because your "rule" does not always hold.

It is always true that $x = y \implies b^x = b^y$ because the function $f_b(x) = b^x$ is well-defined (assume we are working with positive real numbers).

It is not always the case of the converse, $b^x = b^y \implies x = y$. When $f(x) = f(y) \implies x = y$, the function is injective.

Observe for example the function $f_1(x) = 1^x$ is not injective since $1^1 = 1^2$ but $1 \ne 2$. However, for $b \ne \pm 1$ (ignoring the case $b=0$), $f_b(x) = b^x$ is injective (try to prove this).

Answer 6

As a complement of the other answers:

consider the function $f(x)=a^x+b^x$ with $a>0$ and $b>0$. Assume for simplicity that $a\neq 1$ and $b\neq 1$. There are a few cases to consider:

1. If $a>1$ and $b>1$, then $a^x$ and $b^x$ are increasing, so $f(x)=a^x+b^x$ is increasing. As a consequence, $f$ is one-to-one so, for every constant $c$, the equation $f(x)=c$ has at most one solution. Precisely, the range of $f$ is $(0,\infty)$ so it has exactly one solution when $c>0$ and $0$ otherwise.

   This case applies to the equation $2^x+4^x=6$ which has only one solution, $x=1$.

2. Similarly, when $0<a<1$ and $0<b<1$, the function $f(x)=a^x+b^x$ is decreasing with range $(0,\infty)$, so every equation $f(x)=c$ has exactly one solution when $c>0$ and $0$ otherwise.

3. When $a>1$ and $0<b<1$, there are two or zero solutions to $f(x)=c$. The simple case is when $b=\frac{1}{a}$, since in that case, $f$ is even.

   More generally, $f'(x)=\ln(a)a^x + \ln(b)b^x$. Since $\ln(a)>0$, $\ln(b)<0$, $a^x$ is increasing and $b^x$ is decreasing, then $f'$ is increasing. We have $\lim_{x\to\pm\infty} f'(x)=\pm\infty$, so $f$ is decreasing on $(-\infty,\alpha]$, then increasing on $[\alpha,\infty)$, where $\alpha$ is the solution of $f'(x)=0$ (precisely, $\alpha=\ln\left(\frac{b}{a}\right)\ln\left(\frac{\ln(1/b)}{\ln(a)}\right)$).

   So, if $c>f(\alpha)$, then the equation $f(x)=c$ has exactly two solutions.

Answer 7

The term "solve" generally means a systematic method of finding solutions, and showing that they are exhaustive (that is, no further solutions exist). This method fails on both counts.

First, while for the particular example you give, it's not too difficult to see that 1 is a solution, I don't see any way to generalize this to a general method of solving equations of this type. It appears to just be guess and check, and while guess and check can be considered to be a method of solving in the broadest sense, it's not what people usually mean by the term.

Second, this doesn't show that 1 is the only solution. There are cases where you have $x^a+y^a = x^b+y^b$ but $a \neq b$. One situation where that can happen is if $y = x^{-1}$. Then $x^{-a}+y^{-a} = (x^{-1})^a +(y^{-1})^a = y^a+x^a = x^a+y^a$. In fact, your original statement that $(x^a = x^b) \rightarrow (a = b)$ holds only for $x$ positive real and $a, b$ real.