P2.5 - Exponentials and logarithms

Learning objectives

P2.5.1y = ax and its graphy = ax and its graph. a > 0, a ≠ 1.P2.5.2Laws of logarithmsUse log_a(xy) = log_a x + log_a y, log_a(x/y) = log_a x − log_a y, log_a(x^k) = k log_a x, log_a(1/x) = −log_a x and log_a a = 1, for valid positive arguments and a ≠ 1.P2.5.3Solving equations using logarithmsThe solution of equations of the Students may use the change of base formula. form ax = b.

In $y=a^x$ , the variable is in the exponent. The restrictions $a>0$ and $a\ne1$ give a real exponential curve rather than a sign-changing or constant rule.

$y=a^x,\qquad a>0,\ a\ne1$

Base	Left-to-right behaviour	End behaviour
$a>1$	increasing	$y\to0$ as $x\to-\infty$ ; $y\to\infty$ as $x\to\infty$
$0<a<1$	decreasing	$y\to\infty$ as $x\to-\infty$ ; $y\to0$ as $x\to\infty$

Every permitted base gives $a^0=1$ , so the graph crosses the $y$ -axis at $(0,1)$ . Its domain is all real $x$ , its range is $y>0$ , and $y=0$ is a horizontal asymptote: the curve approaches but never meets the $x$ -axis.

Known graph transformations preserve this structure. For $y=3\cdot2^x+4$ , the $y$ -intercept is $(0,7)$ , the horizontal asymptote is $y=4$ , and the curve is increasing above that line.

Do not draw an $x$ -intercept for $y=a^x$ . A vertical shift changes the horizontal asymptote, while $a=1$ would give the excluded constant graph $y=1$ .

A logarithm is an exponent: $\log_a x=y$ means $a^y=x$ . Therefore $a>0$ , $a\ne1$ , and every logarithm argument must be positive.

$\begin{aligned}\log_a(xy)&=\log_a x+\log_a y,\\ \log_a(x/y)&=\log_a x-\log_a y,\\ \log_a(x^k)&=k\log_a x,\\ \log_a(1/x)&=-\log_a x,\\ \log_a a&=1.\end{aligned}$

These laws are the index laws read through the inverse operation: multiplying powers adds exponents, dividing subtracts them, and raising a power multiplies its exponent.

Solve $\log_2(x-1)+\log_2(x-3)=3$ . The domain requires $x>3$ . Combining gives $\log_2[(x-1)(x-3)]=\log_2 8,$ so $(x-1)(x-3)=8$ and $(x-5)(x+1)=0$ . Only $x=5$ satisfies the domain.

The arguments $x$ , $y$ , $xy$ , or $x/y$ must be positive wherever their logarithms appear. Algebra can create candidate roots outside that domain, so check every final value in the original expression.

To solve $a^x=b$ , apply a logarithm to both sides. The power law moves the unknown exponent in front, where ordinary algebra can isolate it.

$a^x=b\quad\Longrightarrow\quad x\log_c a=\log_c b\quad\Longrightarrow\quad x=\frac{\log_c b}{\log_c a}=\log_a b$

This requires $a>0$ , $a\ne1$ , and $b>0$ . The change-of-base formula allows any convenient valid base $c$ , commonly $10$ or $e$ .

For $5^{2x-1}=17$ , take natural logarithms: $(2x-1)\ln5=\ln17.$ Hence $x=\frac12\left(1+\frac{\ln17}{\ln5}\right)\approx1.380.$ Substitution restores $5^{2x-1}=17$ , providing a check.

First isolate the exponential expression, then take logs of both complete sides. Do not write $\log(a^x)=\log a^x$ ambiguously, and keep full calculator precision until the final rounding.

Next Topic

—P2.6 - TrigonometryRead

P2.5 - Exponentials and logarithms

Learning objectives

Read the shape of an exponential graph

Combine logarithms without losing their domain

Use logarithms to release an unknown exponent

Next Topic