How do you solve #ln e^x -ln e^3 = ln e^9#?

1 Answer
Truong-Son N.
Jul 7, 2016

If you use the properties of logarithms, you will need to refer to these:

  1. #lna^b = blna#
  2. #lne = 1#

OR

#lne^z = z#

where #a# and #b# are constants, #e = 2.718281828cdots#, and #z# is either a constant or a variable. Therefore, we can do this:

#lne^x - lne^3 = lne^9#

#lne^x = lne^9 + lne^3#

#xcancel(lne) = 9cancel(lne) + 3cancel(lne)#

#color(blue)(x) = 9 + 3 = color(blue)(12)#

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