The equations are equivalent to; b=ac,‌‌a=bc+3∕2,‌‌a=cb Raise each side of the second equation to the power of c. Then b=bc2+3c∕2, so we must have 1=c2+3c∕2 (since b>1 for the logarithms to make sense). This has solutions c=−2 or c=−1∕2, but we know c>0 so c=1∕2. The remaining equations read a=b2 and a=2−b. Sketch both curves;
There is exactly one positive solution. The answer is (a)