6. Logarithmic and exponential functions

Learning objectives

6.1Use logarithmic and exponential graphs• Know and use simple properties and graphs of logarithmic and exponential functions, including ln x and e^x.• Understand that f(x) = e^x and g(x) = ln x are inverse functions.• Understand the asymptotic nature of logarithmic and exponential graphs and state equations of asymptotes.• Graphs are limited to y = ke^(nx) + a and y = k ln(ax + b), where n, k, a and b are integers.• Series expansions are not required.6.2Use laws of logarithms• Know and use the laws of logarithms, including change of base of logarithms.• Combine logarithmic expressions and convert logarithms between bases, including natural logarithms.6.3Solve exponential equations• Solve equations of the form a^x = b.

$y=e^x$ and $y=\ln x$ are inverse functions. Their graphs are reflections in $y=x$ : swapping every point $(p,q)$ to $(q,p)$ also swaps domain with range and horizontal features with vertical ones.

Graph	Domain	Range	Intercept	Asymptote
$y=e^x$	all real $x$	$y>0$	$(0,1)$	horizontal $y=0$
$y=\ln x$	$x>0$	all real $y$	$(1,0)$	vertical $x=0$

For $y=ke^{nx}+a$ with $k\ne0$ and $n\ne0$ , the horizontal asymptote is $y=a$ and the $y$ -intercept is $(0,k+a)$ ; its range is $y>a$ if $k>0$ and $y<a$ if $k<0$ . For $y=k\ln(ax+b)$ with non-zero $a,k$ , require $ax+b>0$ , the vertical asymptote is $x=-b/a$ , and the $x$ -intercept satisfies $ax+b=1$ .

$y=4e^x+3$ has horizontal asymptote $y=3$ and $y$ -intercept $(0,7)$ . For $y=\ln(4x-3)$ , the domain is $x>3/4$ , the vertical asymptote is $x=3/4$ , and the $x$ -intercept is $(1,0)$ .

An asymptote is found from the transformed input or output, not copied unchanged. The exponential graph never reaches $y=a$ , and the logarithmic graph is undefined where $ax+b\le0$ . Series expansions are outside this syllabus objective.

For a real logarithm, the base must satisfy $a>0$ and $a\ne1$ , and every logarithm argument must be positive. Logarithm laws rewrite multiplication, division and powers; they do not distribute across addition.

Structure	Logarithm form
$uv$	$\log_a(uv)=\log_a u+\log_a v$
$u/v$	$\log_a(u/v)=\log_a u-\log_a v$
$u^r$	$\log_a(u^r)=r\log_a u$
change of base	$\log_a u=\dfrac{\ln u}{\ln a}=\dfrac{\log_b u}{\log_b a}$
reciprocal bases	$\log_a b=1/\log_b a$

$3\lg x-2\lg(y^2)-3=\lg\left(\frac{x^3}{1000y^4}\right),\qquad x>0,\ y\ne0$

The coefficients become powers, and $3=\lg1000$ supplies the constant inside the quotient. Because the original expression contains $\lg(y^2)$ , its real-domain condition is $y\ne0$ ; the final logarithm has the same condition.

Never write $\log(u+v)=\log u+\log v$ . When solving, record the original positive-argument restrictions before combining logs, then reject any candidate that violates them.

Equation after isolation	Best move
$a^{f(x)}=a^{g(x)}$	equate exponents: $f(x)=g(x)$
$a^{f(x)}=b$ with no useful common base	take logarithms: $f(x)\ln a=\ln b$

$a^x=b,\quad a>0,\ a\ne1,\ b>0\quad\Longrightarrow\quad x=\frac{\ln b}{\ln a}$

Solve $4e^{2x-3}=7e^{5-x}$ . Divide first: $e^{3x-8}=7/4$ . Taking natural logarithms gives $3x-8=\ln(7/4)$ , so $x=\dfrac{8+\ln(7/4)}{3}\approx2.85$ .

Isolate the exponential expression before taking logs. Keep the exact logarithmic form until the final line, then round only if requested. Any logarithm base gives the same solution when used consistently.

For a positive base, $a^x$ is always positive, so $a^x=b$ has no real solution when $b\le0$ . Do not equate exponents unless both sides have first been expressed with the same valid base.

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—Cambridge IGCSE Additional Mathematics (0606)Read

6. Logarithmic and exponential functions

Learning objectives

Connect exponential and logarithmic graphs

Use logarithm laws with valid arguments

Solve exponential equations efficiently

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