math?
posted by dj .
Simplify each of the following, if possible. Write your answers in exponential form, a^b
3^4*3^7*3^6
2^80+3*2^80

math? 
MathMate
There 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
In the second problem why do I put a 1 before 2^80?

math? 
bobpursley
you don't have to, but on the second step it makes factorization more apparent.

math? 
Anonymous
thats why the answer not 3^2^280 I am not understanding the answer

math? 
dj
thats why the answer not 3^2^280 I am not understanding the answer
you don't multiply 3*2=6^80 
math? 
MathMate
The 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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