How do you solve #ln x + ln 5 = ln10#?

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Answer 1 · 2015-11-29T18:13:50

#x=2#

Subtract #ln(5)# from both sides:

#ln(x) = ln(10)-ln(5)#

Use the rule #ln(a)-ln(b)=ln(a/b)#:

#ln(x)=ln(10/5)=ln(2)#.

Since the logarithm is injective, #x# must be #2#. If you want to see it in an other way, take the exponential on both sides:

#e^{ln(x)} = e^{ln(2)}#

Since logarithm and exponential are inverse functions, we have that #e^{ln(z)}=z#. So,

#x=2#