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Problem: Find an equation of the straight line that is tangent to $y= 2^x$ and that passes through the point $(1,0)$.

Attempt: Let $(a, 2^a)$ be the point of tangency. Now we have that $y' = 2^x \ln(2)$, which evaluated at the tangency point becomes $y' = 2^a \ln(2)$.

So the general equation of the tangent line is $y = 2^a \ln(2) (x-a) + 2^a$. If this line is to go through $(1,0)$, then we must demand that $0 = 2^a \ln(2) (1-a) + 2^a$. I rewrote this as $-2^a = 2^a \ln(2) (1-a)$, which simplifies to $ -1 = \ln(2) (1-a)$. Now I'm having trouble finding $a$ from this. If I distribute I get $ a \ln(2) = \ln(2) + 1$ or \begin{align*} a = \frac{ \ln(2) + 1}{\ln(2)} \end{align*}

Can I simplify this any further? Because my textbook says the tangent line has equation \begin{align*} y = 2e \ln(2) (x-1) \end{align*} How can I find this answer?

Kamil
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2 Answers2

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To show the answers are the same, we need to show that if $a=\frac{\ln 2+1}{\ln 2}$ then $2^a=2e$.

Note that $a=1+\frac{1}{\ln 2}$, so $$2^a=2\cdot 2^{1/\ln 2}=2\cdot (e^{\ln 2})^{1/\ln 2}=2e.$$

André Nicolas
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I'm sure your answer is correct. There is an alternative way to do it to get that answer. Instead of using that unknown point to get the equation of the line, use $(1,0)$:

$$y=2^a \ln 2(x-1)$$

Now plug $(a,2^a)$ into it: $$2^a=2^a \ln 2(a-1)$$

This gives you $a=\frac{1}{\ln 2}+1$. This is the same as you got. Then $2^a=e^{a\ln 2}=e^{(\frac{1}{\ln 2}+1)\ln2}=2e$. Now using the simpler equation of the line will give you the answer from the book.

KittyL
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