y=2e^(-x) Is it y=ln(2/x) ?
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$\displaystyle y = -\ln\bigg(\frac{x}{2}\bigg)$
Are they the same? If x=2, y=2e^(-2)=0.2706 So for the inverse, if x=0.2706 y=-ln(0.2706/2)=2 and y=ln(2/0.2706)=2 Am I correct in saying that the two functions (y=-ln(x/2) and y=ln(2/x)) are the same?
Yep its the same
Originally Posted by MegaVortex7 Am I correct in saying that the two functions (y=-ln(x/2) and y=ln(2/x)) are the same? i just want to clarify 11rdc11's response recall: $\displaystyle \log_a (x^n) = n \log_a x$ so $\displaystyle - \ln \bigg( \frac x2 \bigg) = \ln \bigg( \frac x2 \bigg)^{-1} = \ln \bigg( \frac 2x \bigg)$
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