Question

Simplify the following exponential expression. Show your work step by step and list the Properties of Exponents used to solve this problem next to your work.
3x^{0}(2x^{3}t^{2})^{4}
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(4x^{7}y^{4})^{2}

Answer 1

Given problem is

$\frac{3x^0(2x^3t^2)^4}{(4x^7y^4)^2}$

distribute outer exponents using formula: $(a^mb^n)^c=a^{mc}b^{nc}$, we get:

$=\frac{3x^0(2^4x^{3\cdot4}t^{2\cdot4})}{4^2x^{7\cdot2}y^{4\cdot2}}$

Simplify exponents:

$=\frac{3x^0(2^4x^{12}t^8)}{4^2x^{14}y^8}$

plug $x^0=1$

$=\frac{3\cdot1(2^4x^{12}t^8)}{4^2x^{14}y^8}$

simplify exponents

$=\frac{3(16x^{12}t^8)}{16x^{14}y^8}$

simplify (closure property)

$=\frac{48x^{12}t^8}{16x^{14}y^8}$

simplify exponent part using formula : $\frac{a^m}{a^n}=a^{\left(m-n\right)}$ we get:

$=\frac{3x^{\left(12-14\right)}t^8}{y^8}$

Simplify exponents:

$=\frac{3x^{\left(-2\right)}t^8}{y^8}$

send term to denominator to avoid negative exponent

$=\frac{3t^8}{x^2y^8}$

Hence final answer is $\frac{3t^8}{x^2y^8}$.

Answer 2

$\frac{3x^0(2x^3t^2)^4}{(4x^7y^4)^2} = \frac{3(1)(2)^4(x^3)^4(t^2)^4}{(4)^2(x^7)^2(y^4)^2}$      Since, $a^0 = 1$  and  $(ab)^m=a^mb^m$

$\frac{3x^0(2x^3t^2)^4}{(4x^7y^4)^2} = \frac{3(16)x^{12}t^{8}}{16x^{14}y^8}$    Since, $(a^b)^c=a^{bc}$

$\frac{3x^0(2x^3t^2)^4}{(4x^7y^4)^2} = \frac{3t^{8}}{x^{14-12}y^8}$

$\frac{3x^0(2x^3t^2)^4}{(4x^7y^4)^2} = \frac{3t^{8}}{x^{2}y^8}$

Thus,

$\frac{3x^0(2x^3t^2)^4}{(4x^7y^4)^2} = \frac{3t^{8}}{x^{2}y^8}$