Question

Anyone know the answer to this algebra problem?

Answer 1

Answer:
`\bold{a=1\qquad b=(1)/(16)\qquad c=(1)/(64)\qquad d=1\qquad e=(4)/(9)\qquad f=(16)/(81)}`

Explanation:

`\begin{array}l\underline{\quad x\quad}&\underline{\quad 4^(-x)\qquad \qquad}\\-1&4^(-(-1))=4^1=4\\\\0&4^(-(0))=4^0=1\\\\2&4^(-(2))=(1)/(4^2)=(1)/(16)\\\\4&4^(-(4))=(1)/(4^4)=(1)/(64)\end{array}`

`\begin{array}l\underline{\quad \bigg x \quad}&\underline{\quad \bigg((2)/(3)\bigg)^x\qquad \qquad}\\\\-1&\bigg((2)/(3)\bigg)^(-1)=(3)/(2)\\\\0&\bigg((2)/(3)\bigg)^(0)=1\\\\2&\bigg((2)/(3)\bigg)^(2)=(2^2)/(3^2)=(4)/(9)\\\\4&\bigg((2)/(3)\bigg)^(4)=(2^4)/(3^4)=(16)/(81)\end{array}`