Question

2e^3x=4e^5x. find x using log. show work please

Answer 1

2
`e^(3x)` = 4
`e^(5x)`

`e^(3x)` = 2
`e^(5x)` divided both sides by 2

ln (
`e^(3x)` ) = ln (2
`e^(5x)` ) applied "ln" to both sides

ln (
`e^(3x)` ) = ln 2 + ln tex] e^{5x} [/tex] ) applied the log product rule

3x = ln 2 + 5x applied "ln e = 1"

-2x = ln 2 subtracted 5x from both sides

x =
`(ln2)/(-2)` divided both sides by "-2"