Many thanks in advance.

- Thread starter Sazzle
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Many thanks in advance.

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Divide both sides by 26, then take the natural logarithm of both sides. There will be one more easy step after that.

If you need additional help, please let us know where you need help; I can't tell whether you just needed a reminder to use logs, or have forgotten all your algebra!

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If I understand correctly it is \(\displaystyle y=26e^{0.6x} \). If that is correct then \(\displaystyle x=\frac{\log(y)-\log(26)}{0.6} \)Hi. I have some data which yields an exponential relationship between metabolic expenditure and body movements. The curve has an equation "y=26e to the power 0.6x". I would like to learn, please, how to extrapolate x, for when I only know y.

Thank you. That’s what I needed. Yes I have forgotten all my algebra and I get nervous with logarithmic equations. But now I get it.solvefor x, given y. (The equation is already an extrapolation from your data.)

Divide both sides by 26, then take the natural logarithm of both sides. There will be one more easy step after that.

If you need additional help, please let us know where you need help; I can't tell whether you just needed a reminder to use logs, or have forgotten all your algebra!

I have also realised if I switch axes when plotting my data I get to extrapolate a much more useful equation for my purposes!!

Thank you, that is the equation I was trying to rearrange with my shaky algebra skills! AmazingIf I understand correctly it is \(\displaystyle y=26e^{0.6x} \). If that is correct then \(\displaystyle x=\frac{\log(y)-\log(26)}{0.6} \)

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\(\displaystyle x=\frac{ln(y)-ln(26)}{0.6}\)

Thank you. Much appreciated\(\displaystyle x=\frac{ln(y)-ln(26)}{0.6}\)