Calculus / exponential

Question

Find values of a and b so that $y = a·b^x$ and the line $y = x + 2$ are tangent at $x = 0$.

I tried to substitute with the zero and it seem that the $B=1$ at all time but what about the $a$?

Answer 1

If you meant to say that the line $y=x+2$ is a tangent to the curve $y=ab^x$ at the point $x=0$ then it solve it by firstly differentiating the curve: $$y'=a\cdot b^x\ln(b).$$ At the point $x=0$ the slope of the curve is given by $$y'(0)=a\cdot b^0 \ln(b)$$ $$=a\cdot \ln(b).$$ Since the line $y=x+2$ has a gradient of $1$, and since it is a tangent to the curve $y=a\cdot b^x$ at $x=0$, then $$a\cdot \ln(b)=1. \space \space \space (1)$$ Also at $x=0$ the curve and the line have a common point, thus $$(0)+2=a\cdot b^0$$ $$\therefore a=2.$$ Substituting this back into equation $(1)$ yields $$2 \ln(b)=1$$ $$\therefore \ln(b)=\frac{1}{2}$$ $$\therefore b=e^{\frac{1}{2}}$$ Thus the equation of the curve is $$y=2\cdot (e^{\frac{1}{2}})^x=2e^{\frac{x}{2}}.$$