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Question Number 13283 by Tinkutara last updated on 17/May/17

If f : A → B given by 3^(f(x)) + 2^(−x) = 4 is a bijection, then find A and B if possible.

$$\mathrm{If}\:{f}\::\:{A}\:\rightarrow\:{B}\:\mathrm{given}\:\mathrm{by}\:\mathrm{3}^{{f}\left({x}\right)} \:+\:\mathrm{2}^{−{x}} \:=\:\mathrm{4}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{bijection},\:\mathrm{then}\:\mathrm{find}\:{A}\:\mathrm{and}\:{B}\:\mathrm{if}\:\mathrm{possible}. \\ $$

Answered by 433 last updated on 17/May/17

f(A)=B 3^(f(x)) =4−2^(−x) f(x)=((ln (4−2^(−x) ))/(ln 3)) A={4−2^(−x) >0}={4>2^(−x) }= {2^2 >2^(−x) }={2>−x}={x>−2}=(−2,+∞) x>−2⇔−x<2⇔0<2^(−x) <4 ⇔−4<−2^(−x) <0⇔0<4−2^(−x) <4 ⇔−∞<ln (4−2^(−x) )<ln 4 ⇔^(ln 3>0) −∞<((ln (4−2^(−x) ))/(ln 3))<((ln 4)/(ln 3)) B=(−∞,((ln 4)/(ln 3)))

$${f}\left({A}\right)={B} \\ $$$$\mathrm{3}^{{f}\left({x}\right)} =\mathrm{4}−\mathrm{2}^{−{x}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{ln}\:\left(\mathrm{4}−\mathrm{2}^{−{x}} \right)}{\mathrm{ln}\:\mathrm{3}} \\ $$$${A}=\left\{\mathrm{4}−\mathrm{2}^{−{x}} >\mathrm{0}\right\}=\left\{\mathrm{4}>\mathrm{2}^{−{x}} \right\}= \\ $$$$\left\{\mathrm{2}^{\mathrm{2}} >\mathrm{2}^{−{x}} \right\}=\left\{\mathrm{2}>−{x}\right\}=\left\{{x}>−\mathrm{2}\right\}=\left(−\mathrm{2},+\infty\right) \\ $$$${x}>−\mathrm{2}\Leftrightarrow−{x}<\mathrm{2}\Leftrightarrow\mathrm{0}<\mathrm{2}^{−{x}} <\mathrm{4} \\ $$$$\Leftrightarrow−\mathrm{4}<−\mathrm{2}^{−{x}} <\mathrm{0}\Leftrightarrow\mathrm{0}<\mathrm{4}−\mathrm{2}^{−{x}} <\mathrm{4} \\ $$$$\Leftrightarrow−\infty<\mathrm{ln}\:\left(\mathrm{4}−\mathrm{2}^{−{x}} \right)<\mathrm{ln}\:\mathrm{4} \\ $$$$\overset{\mathrm{ln}\:\mathrm{3}>\mathrm{0}} {\Leftrightarrow}−\infty<\frac{\mathrm{ln}\:\left(\mathrm{4}−\mathrm{2}^{−{x}} \right)}{\mathrm{ln}\:\mathrm{3}}<\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{ln}\:\mathrm{3}} \\ $$$${B}=\left(−\infty,\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{ln}\:\mathrm{3}}\right) \\ $$$$ \\ $$

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