Question

3^8 a^10 b^-5 c^2 over 3^12 a^7 b^-3 c^-2 when a =4 b= 8 and c= 3

Answer 1

1

Explanation:

`(3^8a^(10)b^(-5)c^2)/(3^(12)a^7b^(-3)c^(-2))\qquad\text{use}\ (x^n)/(x^m)=x^(n-m)\\\\=3^(8-12)a^(10-7)b^(-5-(-3))c^(2-(-2))=3^(-4)a^3b^(-2)c^(4)\\\\\text{substitute:}\ a=4,\ b=8,\ c=3:\\\\(3^(-4))(4^3)(8^(-2))(3^4)=(3^(-4)\cdot3^4)\bigg(2^2\bigg)^3\bigg(2^3\bigg)^(-2)\\\\\text{use}\ (x^n)(x^m)=x^(n+m)\ \text{and}\ \bigg(x^n\bigg)^m=x^(nm)\\\\=(3^(-4+4))\bigg(2^((2)(3))\bigg)\bigg(2^((3)(-2))\bigg)=(3^(0))(2^6)(2^(-6))\\\\\text{use}\ x^(-n)=(1)/(x^n)\ \text{and}\ (x^n)(x^m)=x^(n+m)`

`=(1)\left(2^(6+(-6))\right)=2^0=1\\\\\text{Used}\ a^0=1\ \text{for any real value of}\ a,\ \text{except 0}.`

Answer 2

`\bf \cfrac{3^8a^(10)b^(-5)c^2}{3^(12)a^7b^(-3)c^(-2)}\implies \cfrac{a^(10)a^(-7)c^2c^2}{3^(12)\cdot 3^(-8)b^(-3)b^5}\implies \cfrac{a^(10-7)c^(2+2)}{3^(12-8)b^(-3+5)}\implies \cfrac{a^3c^4}{3^4b^2}~\hfill \begin{cases} a=4\\ b=8\\ c=3 \end{cases}`

`\bf \cfrac{4^3\cdot ~~\begin{matrix} 3^4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} 3^4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\cdot 8^2}\implies \cfrac{~~\begin{matrix} 64 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} 64 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies 1`