How do you evaluate #8^0 x 2^-3#?

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Answer 1 · 2016-04-07T08:52:07

#8^0x2^(-3)=x/2^(3)#

#color(blue)("Explaining about "8^0)#

Note that #8^0=1#

Note that # 2= 2^1#

Suppose we had #4/2=2#. But #4=2^2# so we have #2^2/2^1#

It is allowed that you do this: #2^2/2^1 = 2^(2-1)= 2^1=2#

Suppose we had #2/2=1# this is the same as #2^1/2^1 = 2^(1-1) =2^0 =1#

So anything raise to the power of 0 is one. However, there is a lot of debate about #0^0# so stay clear of that one!

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Given:#" "8^0x2^(-3)#

Known:# 8^0=1#

so #8^0x2^(-3)=x2^(-3)#

But #2^(-3)# is the same as #1/2^3# giving:

#8^0x2^(-3)=x/2^(3)#