Question

How do you evaluate `a(b^x)` given a=-2, b=3, x=1?

Answer 1

See a solution process below:

Explanation:

Substitute:

• `color(red)(-2)` for `color(red)(a)`
• `color(blue)(3)` for color(blue)(b)#
• `color(green)(1)` for `color(green)(x)`

`color(red)(a)(color(blue)(b)^color(green)(x))` becomes:

`color(red)(-2) xx (color(blue)(3)^color(green)(1)) => color(red)(-2) xx 3 => -6`