Tangent Line 617E5F

1. **State the problem:** Find the equation of the tangent line to the curve $y = 2^x$ that passes through the point $(1, 0)$. 2. **Recall the formula for the tangent line:** The tangent line to $y = f(x)$ at $x = a$ is given by: $$y = f(a) + f'(a)(x - a)$$ where $f'(a)$ is the derivative of $f(x)$ at $x = a$. 3. **Find the derivative of $y = 2^x$:** Using the exponential derivative rule, $$f'(x) = 2^x \ln(2)$$ 4. **Set the tangent line to pass through $(1, 0)$:** Let the tangent point be at $x = a$. Then the tangent line is: $$y = 2^a + 2^a \ln(2)(x - a)$$ Since it passes through $(1, 0)$, substitute $x=1$, $y=0$: $$0 = 2^a + 2^a \ln(2)(1 - a)$$ 5. **Simplify the equation:** $$0 = 2^a [1 + \ln(2)(1 - a)]$$ Since $2^a > 0$ for all real $a$, divide both sides by $2^a$: $$0 = 1 + \ln(2)(1 - a)$$ 6. **Solve for $a$:** $$1 + \ln(2)(1 - a) = 0$$ $$\ln(2)(1 - a) = -1$$ $$1 - a = -\frac{1}{\ln(2)}$$ $$a = 1 + \frac{1}{\ln(2)}$$ 7. **Find the slope at $x = a$:** $$m = f'(a) = 2^a \ln(2)$$ 8. **Write the tangent line equation:** $$y = 2^a + m(x - a)$$ Substitute $a$ and $m$: $$y = 2^{1 + \frac{1}{\ln(2)}} + 2^{1 + \frac{1}{\ln(2)}} \ln(2) \left(x - \left(1 + \frac{1}{\ln(2)}\right)\right)$$ **Final answer:** The equation of the tangent line is $$y = 2^{1 + \frac{1}{\ln(2)}} + 2^{1 + \frac{1}{\ln(2)}} \ln(2) \left(x - 1 - \frac{1}{\ln(2)}\right)$$