adding exponents with unknowns with the same base

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How do you add numbers with the same base but with unknown exponents? For example, $2^{x+2} + 2^{x+2}$ I understand that you do $2^x \cdot 2^2$ but I get stuck here. I don't know what to do from here or where to get the working out for the next step.

Thanks

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2 Answers

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An expression of the form $ a^n + a^m $ is already in simplest form.

You may, however, if you wish: decrement an exponent by factoring out an $a$ from the term, or increment an exponent by factoring out $\frac1 a$ from the term. That is, the above expression can be written equivalently as any of the following:

$ (a)(a^{(n-1)}) + (a)(a^{(m-1)}) $

$\frac{a^{(n+1)}} {a} + \bigl( \frac{1} {a} \bigr)(a^{(m+1)})$

$ $
By factoring out $a$ or $\frac1 a$ more than once, you can write the equation as:

$ (a^3)(a^{(n-3)}) + \bigl( \frac {a^{(m+5)})} {a^5} \bigr) $

which we can tell is equivalent to the original expression by using exponent rules:
$ (a^3)(a^{(n-3)}) + \bigl( \frac {a^{(m+5)})} {a^5} \bigr) = a^{ ( n-3 ) +3 } + \bigl( \frac{1} {a^5} \bigr) \bigl( a^{(m+5)} \bigr) = a^n + ( a^{-5} ) ( a^{(m+5)} ) = a^n + a^{ ( m+5 ) -5 } = a^n + a^m $

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One last thing worth mentioning is that you may also factor out common multiples between the terms. Example:

$ a^n + a^m = ( a^4 ) ( a^{n-4} ) + ( a^7 ) ( a^{m-7} ) = a^4 \Bigl( ( a^{n-4} ) + (a^3) ( a^{m-7} ) \Bigr) $

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Ultimately, while re-writing equations in more complicated forms such as in the manner shown here is not usually useful on its own, doing so $\mathit is$ often useful when working with more complex equations or in writing proofs.
If this response has been insufficient in answering your question, I'd recommend re-visiting the basic rules of exponents, starting with how addition and multiplication are related, and then moving to how multiplication and exponents are related.
Best regards~

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I will assume you are talking about adding terms which have a constant base and some linear monic expression in the exponent.

You can do:

$2^{x+2} + 2^{x+2}$

= $2\cdot2^{x+2}$

= $2^1\cdot2^{x+2}$

= $2^{1+(x+2)}$

= $2^{x+3}$

Using these properties.

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