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September 20, 2014

September 20, 2014

Posted by **dj** on Sunday, August 16, 2009 at 4:29pm.

3^4*3^7*3^6

2^80+3*2^80

- math? -
**MathMate**, Sunday, August 16, 2009 at 4:40pmThere are two different cases here.

3^4*3^7*3^6 is a pure multiplicative expression, so the rules of priority and rules of exponents are applicable.

1. exponentiation has priority over multiplication

2. multiplication of two numbers to the same base a raised to exponents x and y yields a result of a^{x+y}.

So

3^4*3^7*3^6

= (3^{4})*(3^{7})*(3^{6})

= 3^{4+7+6}

=3^{17}

The second expression, 2^80+3*2^80, has two terms with a common factor 2^{80}. So the simplification consists in factorization of the expression, namely

2^80+3*2^80

= 1*2^{80}+ 3*2^{80}

= (1+3)2^{80}

= 4*2^{80}

- math? -
**dj**, Sunday, August 16, 2009 at 4:45pmIn the second problem why do I put a 1 before 2^80?

- math? -
**bobpursley**, Sunday, August 16, 2009 at 4:58pmyou dont have to, but on the second step it makes factorization more apparent.

- math? -
**Anonymous**, Sunday, August 16, 2009 at 5:04pmthats why the answer not 3^2^280 I am not understanding the answer

- math? -
**dj**, Sunday, August 16, 2009 at 5:06pmthats why the answer not 3^2^280 I am not understanding the answer

you dont multiply 3*2=6^80

- math? -
**MathMate**, Sunday, August 16, 2009 at 5:39pmThe rules of priority of operations requires us to do exponents first, so

3*2^80 is the same as 3*(2^80). That is to say, we do the exponentiation of 2 to the power of 80 before multiplication.

If we multiply before exponentiation, we are not following the rules of priority of operations.

Then come the rules of exponentiation, some of the basic ones are :

a^{0}= 1 for any value of a≠0

a^{1}= a

a^{2}= a*a

a^{3}= a*a*a, etc.

a^{-1}= 1/a

a^{-2}= 1/(a*a)

a^{-3}= 1/(a*a*a), etc.

a^{x}* a^{y}= a^{x+y}

This rule can be applied only if the two bases (a) are the same.

This also means that we cannot simplify a^{x}*b^{y}for general values of a and b.

Since the expression

3*2^{80}

=3^{1}* 2^{80}

has 3 and 2 as bases for the exponents, we cannot simplify the expression by the rules of exponents.

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