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

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Answer 1 · 2018-06-05T00:51:27

Please see below.

We can first say that

#(h + k)^2 = h^2 + 2hk + k^2#

Therefore

#(h + k)^2 - 2hk = 23hk#

#(h + k)^2 = 25hk#

#sqrt((h + k)^2) = sqrt(25hk)#

#(h + k)/5 = sqrt(hk)#

Take the log of both sides.

#log((h + k)/5) = log(sqrt(hk))#

#log((h + k)/5) = log(hk)^(1/2)#

Now apply #loga^n = nloga#

#log((h + k)/5) = 1/2(log(hk))#

Recall that #log(an) = loga + logn#.

#log((h + k)/5) = 1/2logh + 1/2logk#

#log((h + k)/5) = 1/2(logh + logk)#

As required.

Hopefully this helps!