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Why does this log rule make sense:

$$ \log(x)^{1/2}= \frac{1}{2} \log x?$$

I can see why this makes sense:

$$\log x^4=4\log(x)$$

But I can't really make sense of why this works. Like why does:

$$log_24^3=3 \log_24 $$

This is a problem I am having trouble finding the derivative for:

$$H(z) = \ln \sqrt{\frac{a^2-z^2}{a^2 + z^2}}=\frac{1}{2}\ln \frac{a^2-z^2}{a^2 + z^2}=\frac{1}{2} \ln(a^2 - z^2) - \ln(a^2 + z^2)$$

right?

so:

$$H'(z) = \frac{1}{2} \cdot \frac{1}{a^2 - z^2} \cdot -2z - \frac{-1}{a^2 + z^2} \cdot 2z$$

$$ = \frac{-2z}{2(a^2 - z^2)} - \frac{2z}{a^2 + z^2}$$

Is this right?

amWhy
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Jwan622
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2 Answers2

2

A little mistake in your simplification step

\begin{align} H(z) &= \ln \sqrt{\frac{a^2-z^2}{a^2 + z^2}}\\ &= \frac 12 \ln \left(\frac{a^2-z^2}{a^2 + z^2}\right)\\ &= \frac 12 \left( \ln (a^2-z^2) -\ln( a^2+z^2)\right)\\ &=\frac 12 \ln (a^2-z^2) - \color{red}{\frac 12} \ln( a^2+z^2)\\ \end{align}

Jaideep Khare
  • 19,293
1

From the definition

$$y=\log(x)^{1/2}\iff e^y=(x)^{1/2}\iff e^{2y}=x\iff2y=\log x\iff y=\frac12 \log x$$

We can easily generalize for $$\log_a b^c=c\log_a b$$

Thus the simplification for the calculation of the derivative is correct, pay attention to

$$\frac{1}{2} \cdot ln(a^2 - z^2) - \color{red}{\frac12}\cdot ln(a^2 + z^2)$$

user
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