How do you solve #lnx-ln5=0#?

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Answer 1 · 2017-01-04T04:07:24

#x=5#

Using the property that #a^(log_a(x))=x# and #ln# is the base-#e# logarithm:

#ln(x)-ln(5) = 0#

#=> ln(x) = ln(5)#

#=> e^ln(x) = e^ln(5)#

#:. x = 5#

Answer 2 · 2017-01-04T04:59:40

#lnx-ln5=0#

Add #ln5# to each side:
#lnx=ln5#

#x=5#