Question 790521
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From the problem context, I am going to presume that "log" with an unspecified base is base *[tex \Large e] rather than the more commonly accepted base 10.  You will avoid a great deal of ambiguity in the future if you use *[tex \Large \ln(x)] as the natural logarithm function of *[tex \Large x] meaning *[tex \Large \log_e(x)] and reserve *[tex \Large \log(x)] to mean *[tex \Large \log_{10}(x)].


We wish to prove:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{\ln(x)}\ =\ x]


Assume


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{\ln(x)}\ =\ x]


And use algebra to obtain an identity.


Take the natural log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{\ln(x)}\right)\ =\ \ln(x)]


Use the properties of logarithms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(b)\ =\ 1]


to write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x)\left(\ln(e)\right)\ =\ \ln(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x)\ =\ \ln(x)]


Which is true for all *[tex \Large x] in the domain of *[tex \Large \ln(x)]


Therefore *[tex \LARGE e^{\ln(x)}\ =\ x] is true for all *[tex \Large x] in the domain of *[tex \Large \ln(x)].


Q.E.D.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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