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Is there a rule for adding exponential terms of like bases just like there are rules for multiplying and dividing such terms?

For example we know that:

$x^1 \cdot x^2 = x^{1+2} = x^3$

But what about for addition (or subtract for that matter)?

$x^1 + x^2 = x^?$

If no such pattern exists, why is that?

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5 Answers

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You can factor $x^1+x^2$ into $x(1+x)$, but it is not a power. There is no addition law for powers in the way you posit because positive integer powers equal repeated multiplication, not addition:

$$x^{n+m}=\underbrace{x\cdot x\cdot x\cdots x}_{n+m}=\underbrace{x\cdot x\cdots x}_n\cdot\underbrace{x\cdot x\cdots x}_m=x^{n}\cdot x^m.$$

However, the addition law would work for multiplication by integers, rather than integer powers:

$$(n+m)x=\underbrace{x+ x+ x\cdots x}_{n+m}=\underbrace{x+x\cdots x}_n \,+\, \underbrace{x+x\cdots x}_m =nx+mx.$$

Again, with $n+m$ a positive integer. There are conceptual quandries about repeated operations here that I will not go into, but suffice it to say these repetition formulas hold as valuable cases.

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Observe that $ \log(A+B) = \log(A) + \log \left(\frac{B}{A} + 1 \right) $. Using this fact, one can show that $ \begin{align} \log (x^a + x^b) &= \log(x^a) + \log ( x^{b-a} + 1 ) \\ &= a\log x + \log ( x^{b-a} + 1 ) \\ &= \log x \left( a + \frac{\log ( x^{b-a} + 1 )}{\log x} \right) \\ \Rightarrow x^a + x^b &= x^{\left( a + \frac{\log ( x^{b-a} + 1 )}{\log x} \right)} \\ \text{In your case, with }& a=1, b=2, \text{ we get} \\ x^1 + x^2 &= x^{\left( 1 + \frac{\log ( x + 1 )}{\log x} \right)} \end{align} $

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Hint $\,\ x + x^2\, =\, x^n,\ \ n> 1\:$ has at most $\:n\:$ roots over a field (or domain) so it cannot possibly hold true for all elements in an infinite domain.

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You can say that $x^1+x^2=x(x+1)$ or $x^a+x^b=x^a(1+x^{b-a})$ by the distributive law, but there is no $k$ such that $x^1+x^2=x^k$

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I'm not sure what it's called or who discovered it, I just came across it.

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