# Find the straight line which is tangent to the graph of $y=2^x$ at some point and which passes through the point (1, 0).

I understand this is probably a duplicate question, but the answers on said duplicates don't make sense to me or don't seem to work for me.

I'm being asked to find the point of tangency and the equation for the tangent line.

So far I have that:

• $$y'=ln(2)2^x$$
• There is a point $$P=(a, 2^a)$$ which is where the aforementioned tangent line passes through the graph
• It passes through (1,0)

Using this information I tried to create the equation for the tangent line.

$$0-2^a=ln(2)2^a(1-a)$$

And then I end up with

$$0=2^a(ln(2)(1-a)+1)$$

And this is where I get lost, I can't seem to extrapolate any information about a point of tangency from this.

• Is extrapolate a deliberate play on words? If so, excellent! – Matthew Leingang Nov 19 '19 at 16:27

Note the derivative of $$y=2^x$$ is $$y'=2^x\ln 2$$, which should match

$$2^a\ln 2 = \frac {2^a-0}{a-1}$$

Solve for $$a$$,

$$a = \frac1{\ln2}+1$$

Thus, the slope is $$2^{(\frac1{\ln2}+1)}\ln2$$ and use the point-slope formula for the point $$(1,0)$$ to obtain the line

$$y=2^{(\frac1{\ln2}+1)}\ln2\>(x-1)$$

For real finite $$a,2^a>0$$

$$\implies(a-1)\ln2=1$$

$$a=?$$

Solve the equation for $$a$$ and you are done. What is the idea? It is you assume the point is $$(a,2^a)$$ then you evaluate the slope at this point as you did $$\ln 2 2^a$$ equal the slope as $$\frac{2^a-0}{a-1}$$ then solve that for $$a$$

One could note that the fact that exponential functions fulfill the relationship $$y'=ky$$ means exactly that given the tangent at a point $$(a,2^a)$$ on the graph and the point $$(b,0)$$ where the tangent crosses the $$x$$-axis, $$b-a$$ is constant. So we just have to find this distance for some point and then apply this constant difference to the fact that we want the tangent to cross the $$x$$-axis at $$(1,0)$$.

The tangent at $$(0,1)$$ (that seems like the easiest one) has slope $$\ln 2$$, and thus crosses the $$x$$-axis at $$-\frac1{\ln2}$$. Which thus is the $$x$$-coordinate difference we are after. So the point of tangency we want has $$x$$ coordinate $$1+\frac1{\ln2}$$.