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How do you simplify and write #3^8 xx 3^0 xx 3^1# with positive exponents?

Answer 1

Ethan Adkins

#3^9#

This is the same as #3^(8+0+1) = 3^9#

Note that #3^0=1#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Why is that so?")#

Suppose we had #a^3/a^2# then this is the same as #a^(3-2)=a^1=a#

#(axxaxxa)/(axxa) = a/axxa/axxa= 1xx1xxa=a#

...................................................... Now lets change that a bit.

We know that #a^2/a^2 = (axxa)/(axxa)=a/axxa/a=1xx1=1#

but there is another way of writing #a^2/a^2# and that is #a^(2-2)=a^0#

So a value that is written as #a^0# is 1 no matter what the value of #a# is.

It is argued that #0^0=1#

My professor tolled me this and who am I to argue.

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Answer 2

Ethan Adkins

#3^9#

#3^8xx3^0xx3^1#

#:.=3^((8+0+1))#

#:.=8^9#

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Answer 3

Genesis Allen

To simplify and write (3^8 \times 3^0 \times 3^1) with positive exponents, you add the exponents together since the bases are the same:

[3^8 \times 3^0 \times 3^1 = 3^{8 + 0 + 1} = 3^9]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.