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Bunuel
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Approach:

Simplify the given equation:

\(\frac{8^r}{4^s} =2^t --> 2^3 ^r=2^t*2^2 ^s\)

\(3r = t + 2s\)

\(r = \frac{2s+t}{3}\)

IMO Option C it is!
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Q: 8^r/4^s=2^t

8^r=2^3r and 4^s=2^2s

so we have: 2^3r/2^2s=2^t

for the rules of the exponents--->x^y/x^z=x^(y-z) this means 2^3r/2^2s=2^(3r-2s)

2^(3r-2s)=2^t---->3r-2s=t---->r=(t+2s)/3

hope this helps.
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Bunuel
If \(\frac{8^r}{4^s}=2^t\), then what is r in terms of s and t ?


A. \(s + t + 1\)

B. \(s + t + 5\)

C. \(\frac{2s+t}{3}\)

D. \(\frac{2st}{3}\)

E. \(\frac{s}{2}+\frac{t}{4}\)


Simplifying the equation, we have:

2^(3r)/2^(2s) = 2^t

2^(3r-2s) = 2^t

3r - 2s = t

3r = 2s + t

r = (2s + t)/3

Answer: C
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Bunuel
If \(\frac{8^r}{4^s}=2^t\), then what is r in terms of s and t ?


A. \(s + t + 1\)

B. \(s + t + 5\)

C. \(\frac{2s+t}{3}\)

D. \(\frac{2st}{3}\)

E. \(\frac{s}{2}+\frac{t}{4}\)

If \(\frac{8^r}{4^s}=2^t\), then what is r in terms of s and t ?

\(2^{3r-2s} = 2^t\)
3r -2s = t
\(r = \frac{2s + t}{3}\)

IMO C
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Bunuel
If \(\frac{8^r}{4^s}=2^t\), then what is r in terms of s and t ?


A. \(s + t + 1\)

B. \(s + t + 5\)

C. \(\frac{2s+t}{3}\)

D. \(\frac{2st}{3}\)

E. \(\frac{s}{2}+\frac{t}{4}\)

Just take 8/4=2 which makes r=s=t=1

Only Option C works.
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