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The equations below relate the distinct positive reabrumbers aa, bb, cc, and dd. Which equation correctly expresses bb in terms of dd? \newline{c=2a2 a3b=(c2)d+1\begin{cases} c = 2a^{2} \ a^{3b} = \left(\frac{c}{2}\right)^{d+1} \end{cases}

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Q. The equations below relate the distinct positive reabrumbers aa, bb, cc, and dd. Which equation correctly expresses bb in terms of dd? \newline{c=2a2 a3b=(c2)d+1\begin{cases} c = 2a^{2} \ a^{3b} = \left(\frac{c}{2}\right)^{d+1} \end{cases}
  1. Given equations: Given equations:\newlinec=2a2 c = 2a^2 \newlinea3b=(c2)d+1 a^{3b} = \left(\frac{c}{2}\right)^{d+1}
  2. Substitute and simplify: \newlineStep 22: Substitute c=2a2 c = 2a^2 into the second equation:\newlinea3b=(2a22)d+1 a^{3b} = \left(\frac{2a^2}{2}\right)^{d+1} \newlinea3b=(a2)d+1 a^{3b} = (a^2)^{d+1}
  3. Simplify right side: \newlineStep 33: Simplify the right side:\newlinea3b=a2(d+1) a^{3b} = a^{2(d+1)}
  4. Set exponents equal: \newlineStep 44: Since the bases are the same, set the exponents equal:\newline3b=2(d+1) 3b = 2(d+1)
  5. Solve for b: \newlineStep 55: Solve for b b :\newline3b=2d+2 3b = 2d + 2 \newlineb=2d+23 b = \frac{2d + 2}{3}

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