Solve the inequality (10^x - 25*2^x - 2*5^x + 50)/(5x - x^2 - 4) >= 0.

Step-by-step solution

1. 1

   Split $10^x$ into its prime bases. Because $10=2\cdot 5$, we have $10^x=2^x\cdot 5^x$. The numerator becomes $2^x5^x-25\cdot 2^x-2\cdot 5^x+50$, four terms that are begging to be grouped.

2. 2

   Factor the numerator by grouping. Group as $2^x(5^x-25)-2(5^x-25)$, which factors to $(2^x-2)(5^x-25)$. Its zeros are $2^x=2\Rightarrow x=1$ and $5^x=25\Rightarrow x=2$.

3. 3

   Factor the denominator. $5x-x^2-4=-(x^2-5x+4)=-(x-1)(x-4)=(x-1)(4-x)$. It vanishes at $x=1$ and $x=4$, so both values are excluded from the domain. Note that $x=1$ is a zero of the numerator and of the denominator, so it can never be a solution.

4. 4

   Track the sign of each factor. Numerator: $2^x-2<0$ for $x<1$ and $5^x-25<0$ for $x<2$, so the numerator is positive on $(-\infty,1)$, negative on $(1,2)$, and positive on $(2,\infty)$. Denominator $(x-1)(4-x)$ is negative on $(-\infty,1)$, positive on $(1,4)$, and negative on $(4,\infty)$.

5. 5

   Combine the signs region by region. $(-\infty,1)$: $+/-=-$. $(1,2)$: $-/+=-$. $(2,4)$: $+/+=+$. $(4,\infty)$: $+/-=-$. Only $(2,4)$ makes the quotient positive, and $x=2$ itself gives numerator $0$ with a non-zero denominator, so it is included.

6. 6

   Check a sample point and the endpoint. At $x=3$ the numerator is $1000-200-250+50=600$ and the denominator is $15-9-4=2$, giving $300\ge 0$ as predicted. At $x=2$ the numerator is $100-100-50+50=0$ over a denominator of $2$, so the quotient is exactly $0$ and the non-strict inequality holds.