## Calculus (3rd Edition)

The function $f(x)$ is continuous on $[0 ,1$ with $f(0)=1 \gt 0$ and $f(1)=2-b \lt 0$ Therefore, by the Intermediate Value Theorem, there is a $c \in[0, 1 ]$ such that $f(c)= 2^c-bc =0 \implies 2^c =bc$. So, the equation $2^x =bx$ has a solution $c$ in $[0 ,1]$.
We are given that $f(x)=2^x- b x$ The function $f(x)$ is continuous on $[0 ,1$ with $f(0)=1 \gt 0$ and $f(1)=2-b \lt 0$ Therefore, by the Intermediate Value Theorem, there is a $c \in[0, 1 ]$ such that $f(c)= 2^c-bc =0 \implies 2^c =bc$ . So, the equation $2^x =bx$ has a solution $c$ in $[0 ,1]$.