Question #8a9cf

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Question #8a9cf

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Answer 1 · 2016-07-20T02:07:50

${log 2}^{x} = \frac{p}{3}$ $log2^x=p/3$

Explanation:

If I understand the question properly, we have:
${log 8}^{x} = p$ $log8^x=p$

And we wish to express ${log 2}^{x}$ $log2^x$ in terms of $p$ $p$ .

The first thing we should note is that ${log 8}^{x} = x log 8$ $log8^x=xlog8$ . This follows from the following property of logs:
${log a}^{b} = b log a$ $loga^b=bloga$

Essentially, we can "bring down" the exponent and multiply it by the logarithm. Similarly, using this property on ${log 2}^{x}$ $log2^x$ , we get:
${log 2}^{x} = x log 2$ $log2^x=xlog2$

Our problem is now boiled down to expressing $x log 2$ $xlog2$ (the simplified form of ${log 2}^{x}$ $log2^x$ ) in terms of $p$ $p$ (which is $x log 8$ $xlog8$ ). The central thing to realize here is that $8 = 2^{3}$ $8=2^3$ ; which means $x log 8 = x {log 2}^{3}$ $xlog8=xlog2^3$ . And again using the property described above, $x {log 2}^{3} = 3 x log 2$ $xlog2^3=3xlog2$ .

We have:
$p = x {log 2}^{3} = 3 x log 2$ $p=xlog2^3=3xlog2$

Expressing $x log 2$ $xlog2$ in terms of $p$ $p$ is now drastically easier. If we take the equation $p = 3 x log 2$ $p=3xlog2$ and divide it by $3$ $3$ , we get:
$\frac{p}{3} = x log 2$ $p/3=xlog2$

And voila - we have expressed $x log 2$ $xlog2$ in terms of $p$ $p$ .

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Question #8a9cf

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