Chapter 3, Problem 90RE

Solution

To find: The value of b for which ${\text{log}}_{b}x$ a vertical stretch of $y=\ln x$ . Also find the same for a vertical shrink of $y=\ln x$ .

For vertical stretch: $e^{-1}<b<e$

For vertical shrink: $0<b<e^{-1}\text{ or }e<b$

Given: $y=\ln x$

Concept used:

After vertical stretch equation $y=f\left(x\right)$ becomes $y=af\left(x\right)$ , where $\left|a\right|>1$

For vertical shrink , $0<\left|a\right|<1$

Calculation:

Consider $y=\ln x$ . Let the function is dilated by factor a , such that $y=a\ln x$

It can be written as $\frac{y}{a}=\ln x\text{ or }{e}^{\frac{y}{a}}=x\mathrm{...}(i)$

Consider $y={\log }_{b}x$

  $\Rightarrow b^y=x\mathrm{...}(ii)$

From (i) and (ii), $b^y=e^{\frac{y}{a}}$

  $\Rightarrow b=e^{\frac{1}{a}}$

For vertical stretch $\left|a\right|>1$

  $\begin{array}{l}a>1\text{ or }a<-1\\ \frac{1}{a}<1\text{ or }\frac{1}{a}>-1\\ e^{\frac{1}{a}}<e\text{ or }{e}^{\frac{1}{a}}>e^{-1}\\ b<e\text{ or }b>e^{-1}\\ e^{-1}<b<e\end{array}$

For vertical shrink $0<\left|a\right|<1$

  $\begin{array}{l}-1<a<0\text{ or }0<a<1\\ -1>\frac{1}{a}\text{ >0 or }\frac{1}{a}>1>0\\ e^{-1}>e^{\frac{1}{a}}\text{ >0 or }{e}^{\frac{1}{a}}>e>0\\ e^{-1}>b>0\text{ or }b>e>0\end{array}$

Conclusion:

For vertical stretch: $e^{-1}<b<e$

For vertical shrink: $0<b<e^{-1}\text{ or }e<b$