Question

Do anyone know how to solve this

Answer 1

`\displaystyle {\rm{ \frac{2 * {4}^(0) * {6}^(2) }{ {2}^(4) * {4}^( - 1) * 6 } }}`

• Any base with zero index equal to 1.

`\displaystyle {\rm{ \frac{2 * \bold{1} * {6}^(2) }{ {2}^(4) * {4}^( - 1) * {6}^(1) } }}`

• 6 to the power 2 means the base 6 is multiplied 2 times.

`\displaystyle {\rm{ \frac{2 * 1 * \bold{36}}{ {2}^(4) * {4}^( - 1) * 6 } }}`

• 2 to the power 4 means the base 2 is multiplied 4 times.

`\displaystyle {\rm{ \frac{2 * 1 * 36}{ \bold{16} * {4}^( - 1) * 6 } }}`

`\displaystyle {\rm{ \frac{2 * 1 * 36}{16 * \bold{ (1)/(4)} * 4} }}`

`\left[ \because \: {x}^( - m) = \frac{1}{ {x}^(m) } \right]`

`\displaystyle {\rm{ \frac{2 * 1 * 36}{ \cancel{16^4} * (1)/( \cancel4) * 6 } }}`

`\displaystyle {\rm{ (72)/(24) }}`

`\displaystyle {\rm{ (2 * 2 * 2 * 3 * 3)/(2 * 2 * 2 * 3) }}`

• Cancel the common factors....

`3`

Explanation:

`\frak{ \pink{Seolle...a_(prodite)}}`

Answer 2

`\huge\text{Hey there!}`

`\mathsf{(2*4^0*6^2)/(2^4 * 4^(-1) *6^1)}\\\\\mathsf{= (2*\bf 1*36)/(2^4 * 4^(-1) *6^1)}\\\\\mathsf{= (2* \bf 36)/(2^4 * 4^(-1) *6^1)}\\\\\mathsf{= (\bf 72)/(2^4 * 4^(-1) *6^1)}\\\\\mathsf{= (72)/(\bf 16* (1)/(4)*6)}\\\\\mathsf{= (72)/(\bf 16*6)}\\\\\mathsf{= (72)/(\bf 24)}\\\\\mathsf{= \bf 3}`

`\huge\textbf{Therefore, your answer should be: }`

`\huge\boxed{\mathsf{3}}\huge\checkmark`

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