How do you simplify #((a^-2b^4c^5)/(a^-4b^-4c^3))^2#?

Question

Benjamin Altschul · Accepted Answer

#((a^(-2)b^4c^5)/(a^(-4)b^(-4)c^3))^2=a^4b^16c^4# Let us use the identities #a^(-m)=1/a^m#, #1/a^(-m)=a^m#, #a^mxxa^n=a^(m+n)#, #a^m/a^n=a^(m-n)# and #(a^m)^n=a^(mn)# Hence #((a^(-2)b^4c^5)/(a^(-4)b^(-4)c^3))^2# = #((1/a^2b^4c^5)/(1/a^4xx1/b^4c^3))^2# = #((a^4b^4b^4c^5)/(a^2c^3))^2# = #(a^(4-2)b^(4+4)c^(5-3))^2# = #(a^2b^8c^2)^2# = #a^(2xx2)b^(8xx2)c^(2xx2)# = #a^4b^16c^4#

Emery Allen · Answer

undefined To simplify the expression $ \left(\frac{a^{-2}b^4c^5}{a^{-4}b^{-4}c^3}\right)^2 $, use the properties of exponents:

First, distribute the exponent outside the parentheses to each term inside:

$$ \left(a^{-2}b^4c^5 \cdot a^{4}b^{4}c^{-3}\right)^2 $$

Now, simplify the expression inside the parentheses by multiplying like bases and adding exponents:

$$ a^{-2+4}b^{4+4}c^{5-3} $$

$$ a^{2}b^{8}c^{2} $$

Now, raise each term to the power of 2:

$$ (a^{2})^2(b^{8})^2(c^{2})^2 $$

$$ a^{4}b^{16}c^{4} $$

So, the simplified expression is $ a^{4}b^{16}c^{4} $.

Benjamin Altschul · Answer

undefined To simplify the expression $(\frac{a^{-2}b^4c^5}{a^{-4}b^{-4}c^3})^2$, use the properties of exponents and perform the necessary operations inside the parentheses first.

1. Combine the like terms in the numerator and denominator:

$\frac{a^{-2}b^4c^5}{a^{-4}b^{-4}c^3} = a^{-2-(-4)}b^{4-(-4)}c^{5-3} = a^{2}b^{8}c^{2}$

2. Now, raise the simplified expression to the power of 2:

$(a^{2}b^{8}c^{2})^2 = a^{2*2}b^{8*2}c^{2*2} = a^{4}b^{16}c^{4}$

So, the simplified form of $(\frac{a^{-2}b^4c^5}{a^{-4}b^{-4}c^3})^2$ is $a^4b^{16}c^4$.