If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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Updated on: 15 Jul 2019, 21:37
This question can be a good trap if we assume that \(2\) is 1st power and \(32\) is the 5th power of \(2\). The test writer wants us to fall into this trap. If we hastily equate \((a-b)\) to \(1\) and \((a+b)\) to \(5\), then we will indeed find ourselves there. Being suspicious might keep us from such precipitate assumptions.
FIRST STEP: What we know for sure is \(x^{(a−b)}=2\) and \(x^{(a+b)}=32\). We have three unknowns \(x\), \(a\), and \(b\). We should also note that \(x^{(a+b)}=32\) is the 5th power of \(x^{(a−b)}=2\). So we can increase \(x^{(a−b)}=2\) to its 5th power and subtract it from the other:
So if \((x^{(a−b)})^5=2^5\), then we have \(x^{(5a-5b)}=32\). Next, by subtracting we have \(x^{(5a-5b)}=x^{(a+b)}\). Since x>0, the powers must be equal. \(5a-5b=a+b\), so we have \(b=\frac{2}{3}*a\). Bingo! So III must be true!
SECOND STEP: Can we figure out anything else with the help of III? If \(b=\frac{2}{3}*a\) and \(x^{(a−b)}=2\), then by substituting \(a\) with \(b\) we might prove II too, let's check (that's the way we need to think during the exam). If \(a=\frac{3}{2}*b\), then by substitution we have \(x^{(1.5b - b)}=2\). If \(x^{0.5b}=2\), then \(x^b=4\). Bingo! So II also must be true!
THIRD STEP: Considering II and III, can we now conclude that I also must be true or \(a=3\)? Let's use what we already have. If \(x^b=4\) and \(x^{(a+b)}=32\) or \(x^a*x^b=32\), then after substitution we have \(x^a*4=32\) or \(x^a=8\).
If \(x^a=8\), then a=3? We need to prove that \(a\) is \(3\) or instead \(a\) can be any number. So far we know that \(x^a=8\), \(x^b=4\), and \(\frac{a}{b}=\frac{3}{2}\). Let's try some numbers:
1. If \(a=3\), then \(x^a=8\) or \(x^3=8\) or \(x=2\).
If \(b=2\), then \(x^b=4\) or \(x^2=4\) or \(x=2\). We can see that \(a=3\) when \(x=2\).
2. If \(a=6\), then \(x^a=8\) or \(x^6=8\) or \(x=\sqrt{2}\).
If \(b=4\), then \(x^b=4\) or \(x^4=4\) or \(x=\sqrt{2}\). We can see that \(a=6\) when \(x=\sqrt{2}\).
3. If \(a=90\), then \(x^a=8\) or \(x^{90}=8\) or \(x=\sqrt[30]{2}\).
If \(b=60\), then \(x^b=4\) or \(x^{60}=4\) or \(x=\sqrt[30]{2}\). We can see that \(a=90\) when \(x=\sqrt[30]{2}\)
Conclusion: depending on \(x\), \(a\) can be any positive number, not only \(3\). Thus I can be true, but not must be true.
Hence D
Originally posted by
JonShukhrat on 15 Jul 2019, 09:03.
Last edited by
JonShukhrat on 15 Jul 2019, 21:37, edited 1 time in total.