Solve the inequality 1/(2^x + 21) <= 3/(2^(x+2) - 1).

Step-by-step solution

1. 1

   Substitute to turn it into a rational inequality. Every occurrence of $x$ sits in an exponent with base $2$, so let $t=2^x$ with $t>0$. Because $2^{x+2}=2^x\cdot 2^2=4t$, the inequality becomes $\frac{1}{t+21}\le\frac{3}{4t-1}$ — an ordinary rational inequality in $t$.

2. 2

   Move everything to one side instead of cross-multiplying. Cross-multiplying is unsafe here: $4t-1$ changes sign at $t=\tfrac14$, and multiplying by a negative quantity would flip the inequality. Write $\frac{1}{t+21}-\frac{3}{4t-1}\le 0$ and combine over the common denominator: $\frac{(4t-1)-3(t+21)}{(t+21)(4t-1)}\le 0$.

3. 3

   Simplify the numerator. $(4t-1)-3(t+21)=4t-1-3t-63=t-64$, so the inequality is $\frac{t-64}{(t+21)(4t-1)}\le 0$.

4. 4

   Discard the factor that never changes sign. Since $t=2^x>0$, the factor $t+21>21>0$ always. Dividing by a positive quantity keeps the direction, leaving $\frac{t-64}{4t-1}\le 0$.

5. 5

   Build the sign chart in $t$. The critical values are $t=64$ (numerator zero) and $t=\tfrac14$ (denominator zero, excluded). The quotient is negative exactly between them, so $\tfrac14<t\le 64$; $t=64$ is kept because the inequality is non-strict, $t=\tfrac14$ is not because the fraction is undefined there.

6. 6

   Convert back to $x$. Write the bounds as powers of $2$: $2^{-2}<2^x\le 2^6$. The function $2^x$ is strictly increasing, so the exponents obey the same order and $-2<x\le 6$.

7. 7

   Check the two endpoints. At $x=6$: $\frac{1}{64+21}=\frac{1}{85}$ and $\frac{3}{256-1}=\frac{3}{255}=\frac{1}{85}$, so equality holds and $x=6$ belongs to the solution. At $x=-2$: $2^{x+2}-1=2^0-1=0$, so the right-hand side is undefined and $x=-2$ must be excluded.