$\displaystyle ln (e^{x^2-y^2})$ Anyone able to help me?
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Follow this. $\displaystyle \log_ab^c = c\log_ab$
Use this formula: $\displaystyle ln (a)^b=b \cdot ln(a)$
Another way to do basically what pickslides and Also sprach Zarathustra suggest is to use the fact that ln(x) and $\displaystyle e^x$ are inverse functions: $\displaystyle ln(e^x)= x$.
Originally Posted by orkdork $\displaystyle ln (e^{x^2-y^2})$ Anyone able to help me? If you say what you're meant to be doing with this expression, then the answer to "anyone able to help me?" might be yes.
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