I wonder which rule dictates that e^(-2x+ln(c)) is equal to e^(-2x) * c I know that the logarithm naturalis is the "reverse-function" of the e-function but why isn't it e^(-2x) + c instead?
2026-05-15 09:41:55.1778838115
Question about natural logarithm in the exponent of the e-function
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Note that $$ e^{-2x + \ln(c)} = e^{-2x}e^{ \ln(c)} =e^{-2x}c$$