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How do you solve #2.5 = ln(x)#?

1 Answer

Jul 29, 2015

#x=e^{2.5}approx 12.1825#

Explanation:

#log_{b}(c)=d# is equivalent to #b^{d}=c#

Since #ln(x)=log_{e}(x)#, #ln(x)=2.5# is equivalent to #e^{2.5}=x#

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