Solve the inequality 2^x/(2^x - 3) + (2^x + 1)/(2^x - 2) + 5/(4^x - 5*2^x + 6) <= 0.

Step-by-step solution

1. 1

   Substitute $t=2^x$ and rewrite $4^x$. Since $4^x=(2^2)^x=(2^x)^2=t^2$, the whole expression becomes rational in $t$: $\frac{t}{t-3}+\frac{t+1}{t-2}+\frac{5}{t^2-5t+6}\le 0$, where $t>0$ because $2^x$ is always positive.

2. 2

   Factor the third denominator so all three share it. $t^2-5t+6=(t-2)(t-3)$, which is exactly the product of the first two denominators. The domain therefore excludes $t=2$ and $t=3$.

3. 3

   Combine over $(t-2)(t-3)$. The numerator is $t(t-2)+(t+1)(t-3)+5=(t^2-2t)+(t^2-2t-3)+5=2t^2-4t+2$, so the inequality reads $\frac{2t^2-4t+2}{(t-2)(t-3)}\le 0$.

4. 4

   Recognise the perfect square. $2t^2-4t+2=2(t^2-2t+1)=2(t-1)^2$, which is never negative. That single observation decides the problem: the quotient can only be $\le 0$ when the numerator is zero, or when the denominator is negative.

5. 5

   Handle the two cases. Numerator zero gives $t=1$, and there $(1-2)(1-3)=2\neq 0$, so $t=1$ is a valid solution. Denominator negative means $(t-2)(t-3)<0$, i.e. $2<t<3$. Together: $t=1$ or $2<t<3$.

6. 6

   Return to $x$. $t=1$ means $2^x=1$, so $x=0$. The band $2<2^x<3$ gives $1<x<\log_2 3\approx 1.585$, using that $2^x$ is increasing and $2^1=2$.

7. 7

   Verify the isolated point. At $x=0$ we get $\frac{1}{1-3}+\frac{2}{1-2}+\frac{5}{1-5+6}=-\tfrac12-2+\tfrac52=0$, which satisfies the non-strict inequality, confirming that $x=0$ is a genuine isolated solution rather than a stray root.