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I am having trouble to solve this inequality in specific.

For $f(x)=6-2^x$ and $g(x)=4^x$

Solve in $\mathbb{R}$: $$ (f+g)(x)<6 $$

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  • $\begingroup$ Please include your own effort. $\endgroup$ – S.C.B. Feb 4 '17 at 15:37
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Hint. Let $t=2^x$ and solve $6-t+t^2<6$. Can you take it from here?

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  • $\begingroup$ So, I was doing the variable exchange wrong, Your hint help me solving the rest of the exercise. Later I had to use the only possible value for y => 2^x=1 <=> x=0 The answer is ]-infinite;0[ Thank you for the help :) $\endgroup$ – GiuR Feb 4 '17 at 15:53
  • $\begingroup$ @GiuR Well done! Your final result is correct. $\endgroup$ – Robert Z Feb 4 '17 at 15:55
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\begin{align} (f+g)(x) &< 6 \iff \quad \text{using the definitions}\\ 6 - 2^x + 4^x &< 6 \iff \quad \text{subtracting $6$ from both sides} \\ -2^x + 4^x &< 0 \iff \quad \text{adding $2^x$ to both sides} \\ 4^x &< 2^x \iff \quad \text{splitting $4^x=(2^2)^x = 2^{2x} = (2^x)^2$} \\ 2^x 2^x &< 2^x \iff \quad \text{dividing both sides by $2^x \ne 0$} \\ 2^x &< 1 \iff \quad \text{property of exponential functions $b^x$} \\ x &< 0 \end{align}

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