e^-x, solve for x
Solve for x
0.5y = e^-x
I believe the answer is x = ln(2/y)
Would anyone be able to explain the intermediate steps please?
If I was trying it I would have got -ln0.5y = x
0.5y = e^-x
I believe the answer is x = ln(2/y)
Would anyone be able to explain the intermediate steps please?
If I was trying it I would have got -ln0.5y = x
Original post by TenOfThem
Okay, thank you.
I didn't realise that.
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