Question

Question #3d361

Answer 1

See below

Explanation:

`(2^8*3^-5*6^0)^-2*(3^-2/2^3)^4*2^28`

Simplify the brackets first: (when you have an exponent to a power you multiply them. (For example: `" " (x^2)^3=x^6`)

`(2^8*3^-5*6^0)^-2=2^-16*3^10*6^0`

`(3^-2/2^3)^4=3^-8/2^12`

Remember anything to the `0 " power " = 1` so
`" "2^-16*3^10*6^0`
`=2^-16*3^10*1`
`=2^-16*3^10`

Now combine all of the factors: when bases are multiplied add the exponents (ex. `2^4*2^3=2^12" "`)

`2^-16*3^10*3^-8/2^12 *2^28`

`=(2^28*3^-2)/2^28`

`= 3^2` or `9`

Answer 2

See a solution process below:

Explanation:

First, simplify the `6` term using this rule of exponents:

`a^color(red)(0) = 1`

`(2^8 * 3^-5 * 6^color(red)(0))^-2 * (3^-2/2^3)^4 * 2^28 =>`

`(2^8 * 3^-5 * 1)^-2 * (3^-2/2^3)^4 * 2^28 =>`

`(2^8 * 3^-5)^-2 * (3^-2/2^3)^4 * 2^28`

Next, use this rule for exponents to simply the terms in both sets of parenthesis:

`(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))`

`(2^color(red)(8) * 3^color(red)(-5))^color(blue)(-2) * (3^color(red)(-2)/2^color(red)(3))^color(blue)(4) * 2^28 =>`

`(2^(color(red)(8) * color(blue)(-2)) * 3^(color(red)(-5) * color(blue)(-2))) * (3^(color(red)(-2) * color(blue)(4))/2^(color(red)(3) * color(blue)(4))) * 2^28 =>`

`(2^-16 * 3^10) * (3^-8/2^12) * 2^28`

Then, rewrite the expression as:

`((2^-16 * 2^28)/2^12) * (3^10 * 3^-8)`

Now, use this rule of exponents to multiply the numerator in the `2`s term and to multiply the `3`s term:

`x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))`

`((2^color(red)(-16) * 2^color(blue)(28))/2^12) * (3^color(red)(10) * 3^color(blue)(-8)) =>`

`(2^(color(red)(-16)+color(blue)(28))/2^12) * (3^(color(red)(10)+color(blue)(-8))) =>`

`(2^12/2^12) * (3^2) =>`

`1 * 9 =>`

`9`

Answer 3

`3^2=9`

Explanation:

`(2^8*3^-5*6^0)^-2*(3^-2/2^3)^4*2^28`

While this looks extremely complicated, it is manageable if it is done one step at a time, applying the laws of indices. (exponents).

"When raising a power to a power, multiply the indices."
Use this law to remove the brackets.

`color(blue)((2^8*3^-5*6^0)^-2)xxcolor(red)((3^-2/2^3)^4)xxcolor(green)(2^28)`

`color(blue)(2^-16*3^10*6^0)xxcolor(red)(3^-8/2^12)xxcolor(green)(2^28)`
`color(white)(xxxxxxx)uarr`
"Anything to the power of `0` is equal to 1"`" "rarrcolor(blue)(6^0 =1)`

Get rid of the negative indices: `x^-m=1/x^m`

`(color(blue)(3^10)xxcolor(green)(2^28))/(color(blue)(2^16) xxcolor(red)(3^8xx2^12))`

"If you are multiplying and the bases are the same, ADD the indices:

`=(3^10 xx 2^28)/(3^8 xx2^28)`

`=(3^10 xx cancel(2^28))/(3^8 xxcancel(2^28))`

"If you are dividing and the bases are the same, SUBTRACT the indices:

`=3^2`

`=9`