Using Log Laws and values to compute this compution

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The discussion focuses on using logarithmic properties to compute the value of \( N = \frac{0.009292}{(\sqrt[3]{582400}+14.23)} \) with four-figure accuracy. Participants detail the steps to derive \( \log(N) \) as \( \log(0.009292) - \log(\sqrt[3]{582400} + 14.23) \). The correct answer, confirmed by the book, is \( 9.507 \times 10^{-4} \). The discussion emphasizes the application of logarithmic identities, particularly in simplifying the cube root calculation.

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Using logs to compute the following, to the four-figure accuracy. $\frac{.009292}{(\sqrt[3]{582400}+14.23)}$

Let N=$\frac{.009292}{(\sqrt[3]{582400}+14.23)}$, then
$\log\left({N}\right)=\log\left({\frac{.009292}{(\sqrt[3]{582400}+14.23)}
}\right)$.

Log N=

$\log\left({.09292}\right)-\log\left({\sqrt[3]{582400}+14.23}\right)$

What to do with the logarithm after the subtraction sign?the answer in the back of the book is 9.507*10^(-4)
 
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Cbarker1 said:
Using logs to compute the following, to the four-figure accuracy. $\frac{.009292}{(\sqrt[3]{582400}+14.23)}$

Let N=$\frac{.009292}{(\sqrt[3]{582400}+14.23)}$, then
$\log\left({N}\right)=\log\left({\frac{.009292}{(\sqrt[3]{582400}+14.23)}
}\right)$.

Log N=

$\log\left({.09292}\right)-\log\left({\sqrt[3]{582400}+14.23}\right)$

What to do with the logarithm after the subtraction sign?

You might substitute:
$$\sqrt[3]{582400} = 10^{\log(\sqrt[3]{582400})} = 10^{\frac 1 3 \log 582400}$$
 

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