If $h^{2} + k^{2} = 23 h k$ $h^2+k^2=23hk$ , where h>0, k>0, show that log $\frac{h + k}{5} = \frac{1}{2} (log h + log k)$ $(h+k)/5=1/2 (logh+logk)$ ?

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If $h^{2} + k^{2} = 23 h k$ $h^2+k^2=23hk$ , where h>0, k>0, show that log $\frac{h + k}{5} = \frac{1}{2} (log h + log k)$ $(h+k)/5=1/2 (logh+logk)$ ?

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Answer 1 · 2018-06-05T00:51:27

Please see below.

Explanation:

We can first say that

${(h + k)}^{2} = h^{2} + 2 h k + k^{2}$ $(h + k)^2 = h^2 + 2hk + k^2$

Therefore

${(h + k)}^{2} - 2 h k = 23 h k$ $(h + k)^2 - 2hk = 23hk$

${(h + k)}^{2} = 25 h k$ $(h + k)^2 = 25hk$

$\sqrt{{(h + k)}^{2}} = \sqrt{25 h k}$ $sqrt((h + k)^2) = sqrt(25hk)$

$\frac{h + k}{5} = \sqrt{h k}$ $(h + k)/5 = sqrt(hk)$

Take the log of both sides.

$log (\frac{h + k}{5}) = log (\sqrt{h k})$ $log((h + k)/5) = log(sqrt(hk))$

$log (\frac{h + k}{5}) = {log (h k)}^{\frac{1}{2}}$ $log((h + k)/5) = log(hk)^(1/2)$

Now apply ${log a}^{n} = n log a$ $loga^n = nloga$

$log (\frac{h + k}{5}) = \frac{1}{2} (log (h k))$ $log((h + k)/5) = 1/2(log(hk))$

Recall that $log (a n) = log a + log n$ $log(an) = loga + logn$ .

$log (\frac{h + k}{5}) = \frac{1}{2} log h + \frac{1}{2} log k$ $log((h + k)/5) = 1/2logh + 1/2logk$

$log (\frac{h + k}{5}) = \frac{1}{2} (log h + log k)$ $log((h + k)/5) = 1/2(logh + logk)$

As required.

Hopefully this helps!

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If $h^{2} + k^{2} = 23 h k$ $h^2+k^2=23hk$ , where h>0, k>0, show that log $\frac{h + k}{5} = \frac{1}{2} (log h + log k)$ $(h+k)/5=1/2 (logh+logk)$ ?

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Explanation:

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Explanation:

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If $h^{2} + k^{2} = 23 h k$ $h^2+k^2=23hk$ , where h>0, k>0, show that log $\frac{h + k}{5} = \frac{1}{2} (log h + log k)$ $(h+k)/5=1/2 (logh+logk)$ ?