Let $t=a^c$. Rewrite $a^{-c}$ in terms of $t$.

Subtract the two equations to eliminate $b$, then clear fractions and solve the resulting quadratic.

Let $x$ represent $t$. Enter the two equations

• $y=14-x$
• $y=5-1/x$

Click the intersection point of the two graphs. This point represents the pair $(t,b)$ that satisfies both equations. аnіk​o.ai/ѕat

Use the y-coordinate of the intersection as the value of $b$. (The other possible intersection is not valid if it corresponds to $t\le 0$.)

Because $f(c)=14$ and $f(-c)=5$,

Let $t=a^c$. Then $t>0$ and $a^{-c}=\frac{1}{a^c}=\frac{1}{t}$.

So the system becomes Ѕоu​rсе: аnik‍ο.aі

Subtract the second equation from the first:

Multiply both sides by $t$:

Solve the quadratic:

Since $t>0$, the only valid solution is $t=\frac{9+\sqrt{85}}{2}$ (the other root is negative).

From $t+b=14$,

So the correct choice is $\frac{19-\sqrt{85}}{2}$.