Multiplying with variable and different bases/different exponents

Simonsky

Junior Member
Joined
Jul 4, 2017
Messages
128
Hope I've posted this in the correct category - I'm an older maths learner so not up to date with the educational categories!

I thought I knew the rules for exponents but realised I was not sure what to do with a situation with different BASES and different exponents. so how would I simplify: 10^-1(y^2) or can it be simplified?

Thanks
 
Hope I've posted this in the correct category - I'm an older maths learner so not up to date with the educational categories!

I thought I knew the rules for exponents but realised I was not sure what to do with a situation with different BASES and different exponents. so how would I simplify: 10^-1(y^2) or can it be simplified?

Thanks

Does your problem look like:

10-1y2 → which can be written as → [10^-1](y^2)

or

10-1(y2) → which can be written as → 10^[-1(y^2)]
 
… how would I simplify: (10^-1)(y^2) …
We can use a property of exponents to rewrite the power of 10 without a negative exponent; some people consider that to be a simplification, while others consider the actual product (it turns out to be a simple ratio) to be a simplification.

The following property defines a meaning for negative exponents:

\(\displaystyle x^{-n} = \dfrac{1}{x^n}\)

In other words, we can express the reciprocal of a power by changing the sign on the exponent.

\(\displaystyle \text{32 is a power of 2}\)

\(\displaystyle 2^{5} = 32\)

\(\displaystyle \text{The reciprocal of} \; 32 \; \text{is} \; \dfrac{1}{32}\)

\(\displaystyle 2^{-5} = \dfrac{1}{2^{5}} = \dfrac{1}{32}\)


\(\displaystyle 7^{4} = 2401\)

\(\displaystyle 7^{-4} = \dfrac{1}{7^{4}} = \dfrac{1}{2401}\)


\(\displaystyle \left(\dfrac{1}{3}\right)^{-2} = \dfrac{1}{(\frac{1}{3})^2} = \dfrac{1}{\frac{1}{9}} = 9\)



\(\displaystyle \dfrac{1}{x^{-6}} = \dfrac{1}{\frac{1}{x^6}} = x^6\)
 
Last edited:
It's definitely: (10^-1) (y^2)
In computing A*B, you can replace A (or B) with ANYTHING that equals A and then multiply this new way of writing A by B. Now use the great advice by 4bot.
 
An alternate view.

We have the following property for negative exponents:

\(\displaystyle x^{-n} = \dfrac{1}{x^n}\)

Previously, I wrote about the view where changing the sign of the exponent yields the reciprocal of the original power. Using another property of exponents, we can show an alternate approach: A negative exponent denotes raising the reciprocal of the base, instead of the given base.

\(\displaystyle x^{-n} = \bigg(\dfrac{1}{x}\bigg)^{n}\)


\(\displaystyle 2^{-5} = \left(\dfrac{1}{2}\right)^{5} = \dfrac{1}{32}\)


\(\displaystyle 7^{-4} = \bigg(\dfrac{1}{7}\bigg)^{4} = \dfrac{1}{2401}\)


\(\displaystyle \left(\dfrac{1}{3}\right)^{-2} = \left(\dfrac{3}{1}\right)^2 = 9\)


\(\displaystyle \dfrac{1}{x^{-6}} = \dfrac{1}{\big(\frac{1}{x}\big)^6} = \dfrac{1}{\frac{1}{x^6}} = x^6\)


Each viewpoint is handy. When dealing with negative exponents, either will work; I use the one that comes to mind first, depending on algebraic context. With experience, you'll recognize when one may be a bit handier than the other. :cool:
 
Top