Question

On every exponential curve \(y = {b^x}\)\(\left( {b > 0,b \ne 1} \right)\), there is

exactly one point \(\left( {{x_0},{y_0}} \right)\) at which the tangent line to the curve passes through the origin. Show that in every case, \({y_0} = e\). (Hint: You may wish to use Formula 1.5.10.)

Short answer

It is proved that \({y_0} = e\).

Step-by-step solution

Chain Rule of Derivative

Let \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) be a composition of function.The derivative of this function with respect to \(x\) is:

\(\begin{aligned}F'\left( x \right) &= \frac{d}{{dx}}f\left( {g\left( x \right)} \right)\\ &= f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\end{aligned}\)

So \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\).

Proof of the statement

The given function is \(y = {b^x}\).

Differentiating we get:

\(y' = y\ln b\)

So the equation of tangent line at the point \(\left( {{x_0},{y_0}} \right)\) is:

\(y - {y_0} = {y_0}\ln b\left( {x - {x_0}} \right) \Rightarrow y - {b^{{x_0}}} = {b^{{x_0}}}\ln b\left( {x - {x_0}} \right)\)

Since the tangent line passes through the origin so,

\(\begin{aligned}0 - {b^{{x_0}}} &= {b^{{x_0}}}\ln b\left( {0 - {x_0}} \right)\\ - 1 &= \ln b\left( {0 - {x_0}} \right)\\{x_0} &= \frac{1}{{\ln b}}\end{aligned}\)

So we get:

\(\begin{aligned}{l}{y_0} &= {b^{{x_0}}}\\{y_0} &= {b^{\frac{1}{{\ln b}}}}\\{y_0} &= {e^{\ln {b^{\frac{1}{{\ln b}}}}}}\\{y_0} &= {e^{\frac{1}{{\ln b}} \cdot \ln b}}\\{y_0} &= e\end{aligned}\)

Hence it is proved.