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e^-x=6
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X=-ln6
in equivelant logarithic form this says \[ln(6)=-x\]so \[x=-ln(x)\]
what eliza said, not what i said
\[x=-ln(6)\] is right
Remember that \[e^k = a \iff ln\ a = k\] So in your case you have k = -x and a = 6. Therefore: \[e^{-x} = 6 \iff ln(6) = -x\]\[\therefore x = -ln(6)\]
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