How to show that #2^h=4/5#?

Given that #3(4^h)=4(2^k)# and #9(8^h)=20(4^k)#.

1 Answer
Dean R.
Jun 22, 2018

#4/5#

Explanation:

#3(4^h)=4(2^k)#

#3((2^2)^h) = 4(2^k)#

#3(2^{2h}) = 4(2^k)#

#3/4 = 2^{k-2h}#

#9(8^h)=20(4^k)#

#9(2^{3h}) = 20(2^{2k})#

#9/20 = 2^{2k - 3h}#

Here's the slightly tricky part. We need to make #h# out of a linear combination of the exponents. The easy combination that eliminates #k# works fine:

#h = (2k-3h)-2(k-2h) #

#2^h = {2^{2k-3h}}/(2^{k-2h})^2 = (9/20)/(3/4)^2 = 9/20 cdot 16/9 =4/5 #

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