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$E_1 = Ch_1^\alpha$

How can I solve for the exponent alpha?

Is it just $\frac{\frac{E_1}{C}}{ln(\alpha)}$?

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You get

$$h_1^\alpha = \frac{E_1}{C}$$

Tking the log, you get

$$\alpha \ln(h_1) = \ln \frac{E_1}{C}$$

You can easily get $\alpha$ from here. Your answer is almost right, unfortunately, almost right also means wrong ;)

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    Thanks :D so I did have it right.2012-10-25
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    Why is RHS not $\log \frac{E_1}{C}$?2012-10-25
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    Fixed, I knew I should go to sleep instead of answering questions :)2012-10-25