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12.若 $2^{a}+\log _{2} a=4^{b}+2 \log _{4} b$ ，则（ $)$

Gaokao · Math · previous-year question

  1. A.$a>2 b$
  2. B.$a<2 b$correct
  3. C.$a>b^{2}$
  4. D.$a<b^{2}$

Answer

B. $a<2 b$

Explanation

【详解】 设 $f(x)=2^{x}+\log _{2} x$, 则 $f(x)$ 为增函数, 因为 $2^{a}+\log _{2} a=4^{b}+2 \log _{4} b=2^{2 b}+\log _{2} b$ 所以 $f(a)-f(2 b)=2^{a}+\log _{2} a-\left(2^{2 b}+\log _{2} 2 b\right)=2^{2 b}+\log _{2} b-\left(2^{2 b}+\log _{2} 2 b\right)=\log _{2} \frac{1}{2}=-1<0$, 所以 $f(a)<f(2 b)$, 所以 $a<2 b$. $f(a)-f\left(b^{2}\right)=2^{a}+\log _{2} a-\left(2^{b^{2}}+\log _{2} b^{2}\right)=2^{2 b}+\log _{2} b-\left(2^{b^{2}}+\log _{2} b^{2}\right)=2^{2 b}-2^{b^{2}}-\log _{2} b$, 当 $b=1$ 时, $f(a)-f\left(b^{2}\right)=2>0$, 此时 $f(a)>f\left(b^{2}\right)$, 有 $a>b^{2}$ 当 $b=2$ 时, $f(a)-f\left(b^{2}\right)=-1<0$, 此时

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