# 3.9 Exponential and logarithmic functions  (Page 3/3)

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Note that expanse of logarithmic function "f(x)" is along y – axis on either side of axis, showing that its range is R. On the other hand, the expanse of “x” is limited to positive side of x-axis, showing that domain is positive real number. Further, irrespective of bases, logarithmic plots intersect x-axis at the same point i.e. x = 1 as :

$x={a}^{y}={a}^{0}=1$

## Graphs

We have noted the importance of base “1” for logarithmic function. Base “1” plays an important role for determining nature of logarithmic function. Here, we have drawn plots for two cases : (i) 0<a<1 and (ii) a>1. If the base is greater than zero, but less than “1”, then the logarithm function asymptotes to positive y-axis. Since, "x" is positive number, plot falls only in the right side quadrants. Also, note that ${\mathrm{log}}_{\mathrm{0.1}}0.001=3$ is less than ${\mathrm{log}}_{\mathrm{0.1}}0.01=2$ . Hence, plot begins from a higher value to lower value as "x" increases. Further, ${\mathrm{log}}_{\mathrm{0.1}}10=-1$ . We deduce that plot moves below x-axis for "x" greater than "1".

If the base is greater than “1”, then the logarithm function asymptotes to negative y-axis. Since, "x" is positive number, plot again falls only in the right side quadrants. Also, note that ${\mathrm{log}}_{\mathrm{10}}10=1$ is greater than ${\mathrm{log}}_{\mathrm{10}}100=2$ . Hence, plot begins from a lower value to greater value as "x" increases. Further, ${\mathrm{log}}_{\mathrm{10}}10=1$ . We deduce that plot moves above x-axis for "x" greater than "1".

## Logarithmic identities

Some of the important logarithmic identities are given here without proof. Idea, here, is to simply equip ourselves so that we can work with logarithmic functions in conjunction with itself or with other functions.

${\mathrm{log}}_{a}a=1$

${\mathrm{log}}_{a}1=0$

$x={a}^{y}={a}^{{\mathrm{log}}_{a}x}$

${a}^{{\mathrm{log}}_{b}c}={c}^{{\mathrm{log}}_{b}a}$

${\mathrm{log}}_{a}\left(xy\right)={\mathrm{log}}_{a}\left(x\right)+{\mathrm{log}}_{a}\left(y\right)$

${\mathrm{log}}_{a}\left(\frac{x}{y}\right)={\mathrm{log}}_{a}\left(x\right)-{\mathrm{log}}_{a}\left(y\right)$

${\mathrm{log}}_{a}\left({x}^{y}\right)=y{\mathrm{log}}_{a}\left(x\right)$

${\mathrm{log}}_{a}\left({x}^{\frac{1}{y}}\right)=\frac{{\mathrm{log}}_{a}\left(x\right)}{y}$

## Change of base

Sometimes, we are required to work with logarithmic expressions of different bases. In such cases, we convert them to same base, using following relation :

${\mathrm{log}}_{a}\left(x\right)=\frac{{\mathrm{log}}_{b}\left(x\right)}{{\mathrm{log}}_{b}\left(a\right)}$

Problem : Find the domain of the function given by :

$f\left(x\right)=\frac{1}{{\mathrm{log}}_{10}\left(1-x\right)}$

Solution : The given function is reciprocal of a logarithmic function. Therefore, we first need to ensure that logarithmic function does not evaluate to zero for a value of “x”. A logarithmic function evaluates to zero if its argument is equal to “1”. Hence,

$1-x\ne 1$

$⇒x\ne 0$

Further domain of logarithmic function is a positive real number. For that :

$1-x>0$

$⇒1>x$

$⇒x<1$

Combining two results, the domain of the given function is :

$\text{Domain of “f”}=\left(-\infty ,1\right)-\left\{0\right\}$

Problem : Find all real values of “x” such that :

$1-{e}^{\frac{1}{x}-1}>0$

Solution : Solving for the exponential function, we get following inequality :

$⇒{e}^{\frac{1}{x}-1}<1$

Now, we know that domain of an exponential function is “R”. However, this information is not helpful here to find values of “x” that satisfies the inequality. Taking natural logarithm on either side of the equation,

$⇒\frac{1}{x}-1<{\mathrm{log}}_{e}1$

But, logarithm of “1” i.e. ${\mathrm{log}}_{e}1$ is zero. Hence,

$⇒\frac{1}{x}-1<0$

$⇒\frac{1-x}{x}<0$

This is rational inequality. In order to solve this inequality, we need to multiply each side of the inequality by -1 so that "x" in "1-x" becomes positive. This multiplication changes the inequality to "greater than" inequality.

$⇒\frac{x-1}{x}>0$

Here, critical points are 0 and 1. The rational expression is positive in the intervals on either side of middle interval.

Picking the intervals for which expression is positive :

$\text{Domain}=\left(-\infty ,0\right)\cup \left(1,\infty \right)$

## Logarithmic inequality

The nature of logarithmic function is dependent on the base value. We know that base of a logarithmic function is a positive number excluding “1”. The value of “1” plays important role in deciding nature of logarithmic function and hence that of inequality associated to it. Let us consider an equality :

${\mathrm{log}}_{a}x>y$

What should we conclude : $x>{a}^{y}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x<{a}^{y}$ ? It depends on the value of “a”. We can understand the same by considering LHS of the inequality equal to an exponent z :

${\mathrm{log}}_{a}x=z$

If a>1, then " ${a}^{z}$ " will yield a greater “x” than " ${a}^{y}$ ", because z>y (it is given by the inequality). On the other hand, if 0<a<1, then " ${a}^{z}$ " will yield a smaller “x” than " ${a}^{y}$ ", because z>y. We can understand this conclusion with the help of an example. Let

${\mathrm{log}}_{2}x>3$

Let

${\mathrm{log}}_{2}x=4$

Clearly, ${2}^{4}>{2}^{3}\phantom{\rule{1em}{0ex}}\text{as}\phantom{\rule{1em}{0ex}}16>8$

Let us now consider a<1,

$\mathrm{log}{}_{0.5}x>3$

Let

${\mathrm{log}}_{0.5}x=4$

Clearly, ${0.5}^{4}<{0.5}^{3}\phantom{\rule{1em}{0ex}}\text{as}\phantom{\rule{1em}{0ex}}0.0625<0.125$ . Thus, we finally conclude :

${\mathrm{log}}_{a}x>y⇔x>{a}^{y};\phantom{\rule{1em}{0ex}}a>1$ $\mathrm{log}{}_{a}x>y⇔x<{a}^{y};\phantom{\rule{1em}{0ex}}0

We have used two ways notation to indicate that interpretation of logarithmic inequality is true in either direction. Similarly, we can conclude that :

${\mathrm{log}}_{a}x1$ $\mathrm{log}{}_{a}x{a}^{y};\phantom{\rule{1em}{0ex}}0

Logarithmic inequality involving logarithmic functions on same base on either side of the inequality can be interpreted as :

${\mathrm{log}}_{a}x>{\mathrm{log}}_{a}y⇔x>y;\phantom{\rule{1em}{0ex}}a>1$ ${\mathrm{log}}_{a}x>{\mathrm{log}}_{a}y⇔x

Alternate method

We know that if x>y, then :

${a}^{x}>{a}^{y};\phantom{\rule{1em}{0ex}}a>1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{a}^{x}<{a}^{y};\phantom{\rule{1em}{0ex}}0

Let us now consider the inequality ${\mathrm{log}}_{a}x>y$ . Following above deduction, we use the term of inequality as powers on a common base :

${a}^{{\mathrm{log}}_{a}x}>{a}^{y};\phantom{\rule{1em}{0ex}}a>1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{a}^{{\mathrm{log}}_{a}x}<{a}^{y};\phantom{\rule{1em}{0ex}}0

Using identity, $x={a}^{{\mathrm{log}}_{a}x}$ ,

$x>{a}^{y};\phantom{\rule{1em}{0ex}}a>1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x<{a}^{y};\phantom{\rule{1em}{0ex}}0

Similarly, we can deduce other results of logarithmic inequalities.

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