# Simplify: [(a^ 5/4) (a^7/3)] ^-1/2 how on earth do i solve this? thanks

hala718 | Certified Educator

[(a^5/4)(a^7)/3]^-1/2

First we will separate the terms and distribute the power for all terms:

=[(a^5)(a^7) / (3)(4) ]^-1/2

Now since we have a negative power, we can flip the ratio and substitute with the positive power.

= [(3)(4)/(a^5)(a^7)]^1/2

Now we know that (a^5)*(a^7) = a^(5+7)= a^12

= (12/a^12)^1/2

= [(12)^1/2] / (a^12)^1/2

= sqrt(12) / (a^12)^1/2

But, sqrt(12) = sqrt(2*2*3)= 2*sqrt(3)

and (a^12)^1/2= a^(12*(1/2)= a^(12/2)= a^6

= 2sqrt(3)/ a^6

krishna-agrawala | Student

To solve this problem you need to remember and use two simple rules: These are:

(a^x)(a^y) = a^(x+y)

and

(a^x)^y = a^(x*y)

Using these rules we simplify the given expression as follows:

[(a^5/4)(a^7/3)]^(-1/2)

= [(a^(5/4 + 7/3)]^(-1/2)

= [(a^(15/12 + 28/12)]^(-1/2)

= [(a^(43/12)]^(-1/2)

= [(a^(43/12)]^(-1/2)

= a^[(43/12)*(-1/2)]

= a^(-43/24)

giorgiana1976 | Student

First, we'll group the numerators and denominators,like this:

[(a^5*a^7)/(4*3)]^(-1/2)

If we'll multiply 2 exponentials which have the same base, we'll add the exponents, so :

a^5*a^7 = a^(5+7) = a^12

The expression will become:

(a^12/4*3)^(-1/2)

Now, we'll raise each term of the ratio, numerator and denominator, to the given power:

(a^12)^(-1/2)/(4*3)^(-1/2)

To raise an exponential function to a power, we'll multiply the exponents:

(a^12)^(-1/2)  =a^(-12/2) = a^(-6) = 1/a^6

(4*3)^(-1/2) = 1/(4*3)^1/2 = 1/sqrt4*sqrt3 = 1/2*sqrt3

The expression will become:

(1/a^6)/(1/2*sqrt3) = 2*sqrt3/a^6

neela | Student

To simplify {(a^5/4)(a^7/3)}^(-1/2

Solution:

Mind a^5/4 is different from a^(5/4). A^5/4 = (1/4)a^5 by commutative law.

(a^5/4)(a^7/3)  = {{1/7)a^5}{(1/3)a^7} ] ^-1/2

= (1/4)(1/3)a^(5+7) }^-1/2 , by index law  (a^m)(a^n) = a ^(m+n)

={ (1/12) a^12}^-1/2.

= {1/(28a^12)}/2 , as  a^-1/2 = a^-1 and a^-1/2 is not a^(-1/2) and they are different.

= (1/(12a^12)/2

= 1/(24a^12).