Question

How do you solve `36^(2p)=216^(p-1)`?

Answer 1

`p=-3`

Explanation:

Before solving this exponential equation we can recognize that
`" "`
`color(red)36" and " color(red)216 " "` can be written as powers of base `" "color(red)6`
`" "`
`color(red)(36 = 6^2`
`" "`
`color(red)(216 = 6^3`
`" "`
`36^(2p) = 216^(p-1)`
`" "`
`rArrcolor(red)((6^2))^(2p) = color(red)((6^3))^(p-1)`
`" "`
`rArr6^(4p) = 6^(3(p-1))`
`" "`
Therefore,
`" "`
`4p = 3(p-1)`
`" "`
`rArr 4p = 3p - 3`
`" "`
`rArr 4p - 3p = -3`
`" "`
`rArr p = -3`