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 = $\frac{ln2}{-2}$ divided both sides by "-2"