How do you simplify #(4xx10^(-5))^(-6)# ?

#(4*10^-5)^-6#

Does the raising a power to an exponent rule apply here when you multiply -5 by -6?

Thanks.

2 Answers
Jumbotron
Apr 8, 2018

Yes it does affects the exponent, for example;

Explanation:

#(a^2)^3 = (a^2) (a^2) (a^2) = (a^(2 + 2 + 2)) = a^6#

Same thing as;

#(a^2)^3 = (a^(2 xx 3)) = a^6#

Also recall;

#a^-2 = 1/a^2#

Now we solve the question;

#(4 xx 10^-5)^-6 = (4^-6 xx 10^(-5 xx - 6)) = 4^-6 xx 10^30#

or

#(4 xx 10^-5)^-6 = 1/(4 xx 10^-5)^6 = 1/(4^6 xx 10^(-5 xx 6)) = 1/(4^6 xx 10^-30) = (4^6 xx 10^-30)^-1 = (4^(6 xx - 1) xx 10^(-30 xx -1)) = 4^-6 xx 10^30#

EZ as pi
Apr 9, 2018

#1/(4^6 xx 10^-30) = 10^30/4^6#

Explanation:

The rule of multiplying the indices does apply here.

However, notice that there are twp factors inside the bracket so both of them have to be raised to the power of #-6#

#(4xx10^-5)^-6" "rarr# compare with #(xy^2)^5 = x^5y^10#

#= 4^-6 xx 10^(-5xx-6)#

#= 4^-6 xx 10^30#

#= 10^30/4^6#

You could also answer as # 1/(4 xx 10^-5)^6#

Which leads to #1/(4^6 xx 10^-30)#

Related questions
See all questions in Negative Exponents
Impact of this question
3117 views around the world
You can reuse this answer
Creative Commons License