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$\log N=\frac{1}{2}(\log24-\log0.375-6\log3)$ find value N.

I did it below step

$\log N=\frac{1}{2}(\log64-6\log3)$
$\log N=\frac{1}{2}(\log0.877)$
I don't know how continue further

the answer from book is

Thanks your help!

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You need to learn the rules for adding, subtracting and multiplying logarithms, before you can hope to do this kind of problem. The rules you need are $\log x + \log y = \log xy$, $\log x - \log y = \log (x/y)$ and $ x \log y = \log y^x$. Good luck. – user22805 May 23 '12 at 6:57
up vote 3 down vote accepted

$$\log N=\frac{1}{2}(\log24-\log0.375-6\log3)$$ $$\log N=\frac{1}{2}(6\log2 -6\log3)$$ $$\log N=(3\log2 -3\log3)$$ $$\log N=(\log8 -\log27)$$

Basically, $$a \log b = \log (b^a)$$

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Basically $8 = 2^3$ and $27 = 3^3$. – Eugene May 23 '12 at 6:02

$\log N=\frac{1}{2}(\log24-\log0.375-6\log3)$

= $\frac{1}{2}(\log24-(\log0.375+\log3^6))$ since b $\log a = log a^b$

= $\frac{1}{2}(\log24-(\log0.375 *3^6))$

=$\frac{1}{2}\log\frac{24}{0.375 *3^6}$

= $\frac{1}{2}\log\frac{8*3}{(0.5)^3*3 *3^6}$

=$\frac{1}{2}\log\frac{8}{(\frac12)^3 *3^6}$

= $\frac{1}{2}\log\frac{8*8}{3^6}$ =$\frac{1}{2}\log(\frac{8}{3^3})^2$


There fore, $N = \frac{8}{27}$

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$$\log N=\frac{1}{2}(\log24-\log0.375-6\log3)$$

$$\log N= \log \left(\sqrt{\frac{24}{\frac{3}{8}\times 3^6} }\right)$$

$$ N= \sqrt{\frac{3 \times 8 \times 8}{3 \times 3^6} } = \frac{8}{27} = \left(\frac23\right)^3 $$

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Note that logarith is 1-1 function so you can equate arguments of two logarithms of the same base provided the values are identical. Also try to use suggestion given by Nunoxic.

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