cookies Posted May 21, 2007 Posted May 21, 2007 Okay, I have a few questions about putting exponents in simplest form .. Example 5^5 x 5^5 = 5^10 just add the exponents 9^5 x 8^5 = 72^10?? I understand how to do it if its the same number but i cant understand if its diff please tell me..
imp Posted May 21, 2007 Posted May 21, 2007 Okay, I have a few questions about putting exponents in simplest form .. Example 5^5 x 5^5 = 5^10 just add the exponents 9^5 x 8^5 = 72^10?? I understand how to do it if its the same number but i cant understand if its diff please tell me.. Cookies, the number being raised to a power is called the "base". To multiply two or more bases raised to some power, we add the exponents, keeping the base the same. Two non-identical bases raised to some power CANNOT have their exponents added. FYI, similarly, two like bases raised to some power may be DIVIDED by subtracting their exponents. Imp.
Bignose Posted May 21, 2007 Posted May 21, 2007 Well, firstly, you should have some software that can compute these numbers for you. Both Excel and Windows calculator can calculate these numbers ... 9^5 + 8^5 = 1934927632 whereas 72^10 = 3743906242624487424, clearly not the same. However, 72^5 = 1934927632. If and only if the exponents are the same, you can mulitply the bases: 9^5 * 8^5 = (9*8)^5 = 72^5. But, if it is 9^4 + 8^5, there is no simplification possible.
Tom Mattson Posted May 21, 2007 Posted May 21, 2007 So how would i do the 1 with diff base? You can do it like this. [math]a^mb^m=(a \cdot a \cdot\cdot\cdot a)(b \cdot b \cdot\cdot\cdot b)[/math], where each of the parentheses contains m factors. You can then use the commutative and associative properties of multiplication to rearrange the various factors of a and b as follows. [math](ab)\cdot(ab)\cdot\cdot\cdot(ab)[/math] Now you have m factors of (ab). But this is precisely (ab)m. So we arrive at the rule: [math]a^mb^m=(ab)^m[/math] So back to your example... [math]9^5 \cdot 8^5=(9 \cdot 8)^5=72^5[/math]
the tree Posted May 21, 2007 Posted May 21, 2007 Using Tom's method, you can also note that amb(m+p)=(ab)mbp and probably lots of other noteworthy identities.
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