Solve the inequality (14^x - 7^(x+1) - 2^(x+1) + 14)/(-x^2 + 4x) >= 0.

Step-by-step solution

1. 1

   Rewrite each exponential in base $2$ and base $7$. Since $14=2\cdot 7$, $14^x=2^x7^x$; also $7^{x+1}=7\cdot 7^x$ and $2^{x+1}=2\cdot 2^x$. The numerator becomes $2^x7^x-7\cdot 7^x-2\cdot 2^x+14$.

2. 2

   Group the four terms into a product. $7^x(2^x-7)-2(2^x-7)=(7^x-2)(2^x-7)$. The zeros are $7^x=2\Rightarrow x=\log_7 2\approx 0.356$ and $2^x=7\Rightarrow x=\log_2 7\approx 2.807$.

3. 3

   Factor the denominator. $-x^2+4x=x(4-x)$, which is zero at $x=0$ and $x=4$; both are excluded from the domain. Ordering all four critical values gives $0<\log_7 2<\log_2 7<4$.

4. 4

   Sign of the numerator. $(7^x-2)$ is negative for $x<\log_7 2$ and positive after; $(2^x-7)$ is negative for $x<\log_2 7$ and positive after. So the product is positive on $(-\infty,\log_7 2)$, negative on $(\log_7 2,\log_2 7)$, and positive on $(\log_2 7,\infty)$.

5. 5

   Sign of the denominator. $x(4-x)$ is a downward parabola with roots $0$ and $4$: negative outside $[0,4]$ and positive strictly between them.

6. 6

   Combine and pick the regions where the quotient is non-negative. $(-\infty,0)$: $+/-=-$. $(0,\log_7 2)$: $+/+=+$. $(\log_7 2,\log_2 7)$: $-/+=-$. $(\log_2 7,4)$: $+/+=+$. $(4,\infty)$: $+/-=-$. The two numerator zeros are included because the inequality is non-strict and the denominator is non-zero there.

7. 7

   Spot-check one point in each accepted region. At $x=0.2$ the numerator is $14^{0.2}-7^{1.2}-2^{1.2}+14\approx 3.067$ over a denominator of $0.76$, giving about $4.04>0$. At $x=3$ the numerator is $2744-2401-16+14=341$ over $-9+12=3$, giving $\tfrac{341}{3}\approx 113.7>0$.