Question

How do you solve `2ln3x+1=5`?

Answer 1

`x = e^2/3`

Explanation:

`" "`
`2ln3x + 1 = 5`
`" "`
`rArr 2ln3x = 5-1`
`" "`
`rArr2ln3x = 4`
`" "`
`rArrln3x = 4/2`
`" "`
`rArrln3x = 2`
`" "`
`rArre^(ln3x) = e^2`
`" "`
`rArr 3x = e^2`
`" "`
`rArrx = e^2/3`