Question

Let \(c\) be the \(x\)-intercept of the tangent line to the curve \(y = {b^x}\) \(\left( {b > 0,b \ne 1} \right)\) at the point \(\left( {a,{b^a}} \right)\). Show that the distance between the points \(\left( {a,0} \right)\) and \(\left( {c,0} \right)\) is the same for all values of \(a\).

Short answer

Distance between the points \(\left( {a,0} \right)\) and \(\left( {b,0} \right)\) is the same for all values of \(a\).

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:

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

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

Proof of the given statement

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

Differentiating we get:

\(\begin{aligned}y = {b^x}\\\ln y &= x\ln b\\\frac{1}{y}y' &= \ln b\\y' &= y\ln b\end{aligned}\)

So the equation of tangent line at the point \(\left( {a,{b^a}} \right)\) is \(y - {b^a} = {b^a}\ln b\left( {x - a} \right)\).

Since it is given that \(c\)is an \(x\)-intercept of the tangent line so the equation becomes:

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

Hence distance between the points \(\left( {a,0} \right)\) and \(\left( {c,0} \right)\) is \(\frac{1}{{\left| {\ln b} \right|}}\) for any values of \(a\). Since \(\frac{1}{{\left| {\ln b} \right|}}\) is constant.