For the curve $y=e^{bx}$ you have $y\,'=be^{bx}$; the line $y=10x$ has slope $10$, so it can be tangent to the exponential only at a point where $be^{bx}=10$. Of course this point must lie on the exponential and the straight line, so $10x=e^{bx}$. Substitute $be^{bx}$ for $10$ in this last equation to get $bxe^{bx}=e^{bx}$; dividing by $e^{bx}$ shows that $bx=1$, so $e^{bx}=e$, and $b=\dfrac{10}{e^{bx}}=\dfrac{10}e$.
Alternatively, multiply $10x=e^{bx}$ by $b$ to get $10bx=be^{bx}=10$, deduce that $bx=1$, and proceed as above.
For the curve $y=e^{bx}$ you have $y\,'=be^{bx}$; the line $y=10x$ has slope $10$, so it can be tangent to the exponential only at a point where $be^{bx}=10$. Of course this point must lie on the exponential and the straight line, so $10x=e^{bx}$. Substitute $be^{bx}$ for $10$ in this last equation to get $bxe^{bx}=e^{bx}$; dividing by $e^{bx}$ shows that $bx=1$, so $e^{bx}=e$, and $b=\dfrac{10}{e^{bx}}=\dfrac{10}e$.
Alternatively, multiply $10x=e^{bx}$ by $b$ to get $10bx=be^{bx}=10$, deduce that $bx=1$, and proceed as above.