2
$\begingroup$

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?

  • 0
    @YvesDaoust Haha like really. !!! 2018-02-27

5 Answers 5

3

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.

  • 0
    there is some kind of rule though. $ 2^4 + 2^2 = 20 $ $ log_2(20) = 4.3219280949 $ $ 20 - 2^4 = 4 $ $ 2^.3219280949 = 1.25 = 5/4 $2014-10-07
2

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.

  • 0
    @YvesDaoust It differs only in that there is a value $\,n = 3\,$ where it becomes an equality of polynomials, i.e. their difference is the zero polynomial, which has *infinitely* many roots (every element) in an infinite domain. This cannot occur in the case in the answer.2016-05-09
2

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} $

0

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 $x^1+x^2=x^k$

  • 0
    Certainly there is such a $k$: $k = \log_x (x^1 + x^2)$. It's just that $k$ depends on $x$ and is not very simple.2012-03-15
0

formula

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

  • 0
    Seriously? Its not self-explanatory? It's not a unique result and I'd expect most readers here to be able to verify it for themselves. Just... do it. $x^n + y^m = x^m + x^{\log_x y}=x^m(1 + x^{\frac {\log y}{\log x}})=....$ etc.2016-11-27