Science:Math Exam Resources/Courses/MATH104/December 2012/Question 01 (o)/Solution 1

Define the function $f (x)$ as

f (x) = 5^{x} - 10 x - 7 .

Notice that this function is continuous for all values of $x$ . Hence, we need only find a closed interval where the sign of $f (x)$ changes and invoke the intermediate value theorem to get the desired conclusion.

Consider $x = 2$ and $x = 3$ . If we take these two points and plug them into $f (x)$ , we get

\begin{aligned} f (2) & = 5^{2} - 10 (2) - 7 = - 2 < 0 \\ f (3) & = 5^{3} - 10 (3) - 7 = 88 > 0 \end{aligned}

and so the sign of $f (x)$ changes over the closed interval $[2, 3]$ .

Therefore, there exists a value $c$ in $[2, 3]$ such that

f (c) = 5^{c} - 10 c - 7 = 0

by the Intermediate Value Theorem.