How do you evaluate $8^{0} x 2^{- 3}$ $8^0 x 2^-3$ ?

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Answer 1 · 2016-04-07T08:52:07

$8^{0} x 2^{- 3} = \frac{x}{2^{3}}$ $8^0x2^(-3)=x/2^(3)$

$Explaining about 8^{0}$ $color(blue)("Explaining about "8^0)$

Note that $8^{0} = 1$ $8^0=1$

Note that $2 = 2^{1}$ $2= 2^1$

Suppose we had $\frac{4}{2} = 2$ $4/2=2$ . But $4 = 2^{2}$ $4=2^2$ so we have $\frac{2^{2}}{2^{1}}$ $2^2/2^1$

It is allowed that you do this: $\frac{2^{2}}{2^{1}} = 2^{2 - 1} = 2^{1} = 2$ $2^2/2^1 = 2^(2-1)= 2^1=2$

Suppose we had $\frac{2}{2} = 1$ $2/2=1$ this is the same as $\frac{2^{1}}{2^{1}} = 2^{1 - 1} = 2^{0} = 1$ $2^1/2^1 = 2^(1-1) =2^0 =1$

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

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Given: $8^{0} x 2^{- 3}$ $" "8^0x2^(-3)$

Known: $8^{0} = 1$ $8^0=1$

so $8^{0} x 2^{- 3} = x 2^{- 3}$ $8^0x2^(-3)=x2^(-3)$

But $2^{- 3}$ $2^(-3)$ is the same as $\frac{1}{2^{3}}$ $1/2^3$ giving:

$8^{0} x 2^{- 3} = \frac{x}{2^{3}}$ $8^0x2^(-3)=x/2^(3)$

How do you evaluate 8^0 x 2^-380x2−38^0 x 2^-3?