Question

A student claims that 8^3 •8^-5 is greater than 1. Explain whether the student is correct or not

Answer 1

Will give medal

Answer 2

\[8^3 = 8*8*8 * \frac{ 1 }{ 8*8*8*8*8 }\]

Answer 3

\[= \frac{ 1 }{ 8*8 } = \frac{ 1 }{ 8^2 } = \frac{ 1 }{ 64 }\]

Answer 4

\[\frac{ 1 }{ 64 } = 0.015625..\]

Answer 5

So, the Student is wrong because 0.015625 is not greater than 1

Answer 6

I imagine that the simplest way of going about proving this is to use the Laws of Indices. These are a set of rules which allow us to write something in a different fashion when dealing with an expression with one or more values raised to a certain power. Here's a link which may be of interest to you which lists all of these rules:

http://mathematics.laerd.com/maths/indices-intro.php

The one we'll be concerned here with is to do with the fact that:
\[A ^{n}.A ^{m} = A ^{m + n}\]

So, we can apply this to 8^3.8^-5 to leave us with 8^(3-5), or 8^-2.

Another rule of indices is that:
\[A ^{-m} = \frac{ 1 }{ A ^{m} }\]

and so we can write 8^-2 as 1/8^2. Clearly, 1 divided by any number greater than 1 (i.e. a fraction of 1) will be less than 1 and this should hopefully give you the answer!