Logarithmic and exponential functions

Logarithms and indices Paper 2

A logarithm answers the question "what power?". For a base $a>0$ with $a\ne 1$,
$${\log }_{a}x=y⟺a^y=x$$
So ${\log }_{2}32=5$ because $2^5=32$, ${\log }_{10}0.001=-3$ because ${10}^{-3}=0.001$, and ${\log }_{9}3=\frac{1}{2}$ because $9^{\frac{1}{2}}=3$.

Because $a^y$ is always positive, ${\log }_{a}x$ is only defined for $x>0$. Two values come straight from the definition: ${\log }_{a}1=0$ (since $a^0=1$) and ${\log }_{a}a=1$. Applying the power and then the logarithm undoes itself: ${\log }_a(a^x)=x$ and $a^{{\log }_{a}x}=x$.

Logarithms to base 10 are written $\lg x$ on some papers, and logarithms to base $\mathrm{e}$ are written $\ln x$ (see below).

To find a logarithm without a calculator, write both numbers as powers of the same base: ${\log }_{8}4$ asks for $y$ with $8^y=4$, that is $2^{3y}=2^2$, so $y=\frac{2}{3}$.

The laws of logarithms Paper 2

The index laws $a^m\times a^n=a^{m+n}$, $a^m÷a^n=a^{m-n}$ and $(a^m{)}^k=a^{mk}$ turn into three laws of logarithms:

• ${\log }_a(xy)={\log }_{a}x+{\log }_{a}y$;
• ${\log }_a(\frac{x}{y})={\log }_{a}x-{\log }_{a}y$;
• ${\log }_a(x^k)=k{\log }_{a}x$.

Why the first law works. Let ${\log }_{a}x=m$ and ${\log }_{a}y=n$, so $x=a^m$ and $y=a^n$. Then $xy=a^{m+n}$, so ${\log }_a(xy)=m+n$. The other two follow in the same way.

Two consequences are used often: ${\log }_a(\frac{1}{x})=-{\log }_{a}x$, and ${\log }_a\sqrt{x}=\frac{1}{2}{\log }_{a}x$.

There are no laws for ${\log }_a(x+y)$ or for $\frac{{\log }_{a}x}{{\log }_{a}y}$: these do not simplify. All the logarithms in a law must have the same base (changing the base is not required in this paper).

Solving a logarithmic equation. Use the laws to collect each side into a single logarithm, then remove the logarithms: ${\log }_a\mathrm{f}(x)={\log }_a\mathrm{g}(x)$ gives $\mathrm{f}(x)=\mathrm{g}(x)$, and ${\log }_a\mathrm{f}(x)=c$ gives $\mathrm{f}(x)=a^c$. Finally, check every answer in the original equation: a value that makes any logarithm's argument zero or negative must be rejected.

The exponential function Paper 2

The number $\mathrm{e}=2.71828\dots$ is a special base: the curve $y={\mathrm{e}}^x$ has gradient 1 where it crosses the $y$-axis, and in fact its gradient at every point equals its height. You will use that in differentiation; here the important facts are about its graph.

$y={\mathrm{e}}^x$ is called the exponential function. Its graph:

• passes through $(0,1)$, since ${\mathrm{e}}^0=1$;
• is always above the $x$-axis: ${\mathrm{e}}^x>0$ for every $x$, so ${\mathrm{e}}^x=-2$ has no solution;
• increases everywhere, more and more steeply;
• approaches the $x$-axis as $x\to -\infty$, so $y=0$ is an asymptote.

The graph of $y={\mathrm{e}}^{kx}$. For $k>0$ it has the same shape as ${\mathrm{e}}^x$, rising more steeply when $k$ is larger: this models growth. For $k<0$ it is the reflection of a growth curve in the $y$-axis: it falls from left to right towards the asymptote $y=0$, modelling decay. Every one of these curves passes through $(0,1)$.

A curve such as $y=4-3{\mathrm{e}}^{-2x}$ is a transformation of one of these: as $x\to \infty$, ${\mathrm{e}}^{-2x}\to 0$ and $y\to 4$, so its asymptote is $y=4$, and it crosses the $y$-axis at $4-3=1$.

${\mathrm{e}}^{kx}$ with $k>0$ is a growth curve, steeper for larger $k$; with $k<0$ it is a decay curve. All pass through $(0,1)$ with the asymptote $y=0$.

The natural logarithm Paper 2

The logarithm to base $\mathrm{e}$ is the natural logarithm, written $\ln x$:
$$\ln x=y⟺{\mathrm{e}}^y=x$$
So $\ln x$ and ${\mathrm{e}}^x$ are inverse functions: each undoes the other, $\ln ({\mathrm{e}}^x)=x$ for all $x$ and ${\mathrm{e}}^{\ln x}=x$ for $x>0$. As with any inverse pair, the graph of $y=\ln x$ is the reflection of $y={\mathrm{e}}^x$ in the line $y=x$. So it:

• passes through $(1,0)$, since $\ln 1=0$, and $\ln \mathrm{e}=1$;
• exists only for $x>0$, with the $y$-axis as an asymptote;
• increases everywhere, but more and more slowly.

The laws of logarithms hold for $\ln$ exactly as for any base. They give the standard method for an equation in $\mathrm{e}$: isolate the exponential, then take $\ln$ of both sides. For $\ln$ of an expression, isolate the logarithm and then take $\mathrm{e}$ to the power of both sides.

An equation such as $3{\mathrm{e}}^{2x}-10{\mathrm{e}}^x+3=0$ is a quadratic in disguise: since ${\mathrm{e}}^{2x}=({\mathrm{e}}^x{)}^2$, put $u={\mathrm{e}}^x$. Any negative value of $u$ is rejected, because ${\mathrm{e}}^x>0$.

$y=\ln x$ is the reflection of $y={\mathrm{e}}^x$ in $y=x$: the asymptote $y=0$ of one becomes the asymptote $x=0$ of the other.

Equations with the unknown in the index Paper 2

When the unknown is in an index and the bases cannot be matched, take logarithms of both sides (any base will do; $\ln$ is usual) and use the power law to bring the index down. The equation becomes linear in $x$.

For $3^{2x}=4^{x+1}$:

1. take $\ln$: $2x\ln 3=(x+1)\ln 4$;
2. expand and collect the $x$ terms: $x(2\ln 3-\ln 4)=\ln 4$;
3. divide: $x=\frac{\ln 4}{2\ln 3-\ln 4}=\frac{\ln 4}{\ln \frac{9}{4}}=1.71$ (3 s.f.).

Keep the logarithms exact until the last line, and only then use the calculator. Do not "cancel" logarithms: $\frac{\ln 4}{\ln 2.25}$ is not $\ln \frac{4}{2.25}$.

Inequalities with the unknown in the index Paper 2

Logarithms keep the order of positive numbers ($\ln$ is an increasing function), so taking $\ln$ of both sides of an inequality between positive quantities keeps the inequality sign. The danger comes next: when you divide by a logarithm, check its sign. $\ln 0.8$, and the logarithm of any number between 0 and 1, is negative, and dividing by a negative number reverses the inequality.

For $2^x<5$: $x\ln 2<\ln 5$, and $\ln 2>0$, so $x<\frac{\ln 5}{\ln 2}$.

For ${0.8}^n<0.03$: $n\ln 0.8<\ln 0.03$. Since $\ln 0.8<0$, dividing reverses the sign: $n>\frac{\ln 0.03}{\ln 0.8}=15.7$. If $n$ must be a whole number, the smallest is 16: round up, because 15 does not satisfy the inequality.

These are the calculations behind compound interest, depreciation and decay: "after how many years is the value first above (or below) a target?"

Reducing a relationship to linear form Paper 2

Experimental data often follow $y=k{x}^n$ or $y=k(a^x)$ with unknown constants. A curve is hard to read constants from, but a straight line is easy: its gradient and intercept can be measured. Logarithms turn both relationships into straight lines of the form $Y=mX+c$.

Power law, $y=k{x}^n$. Take $\ln$ of both sides:
$$\ln y=\ln k+n\ln x$$
This is linear in $\ln x$ and $\ln y$: plot $Y=\ln y$ against $X=\ln x$. The gradient is $n$ and the $Y$-intercept is $\ln k$.

Exponential law, $y=k(a^x)$. Take $\ln$:
$$\ln y=\ln k+x\ln a$$
This is linear in $x$ and $\ln y$: plot $\ln y$ against $x$ itself. The gradient is $\ln a$ and the intercept is $\ln k$.

To recover the constants, remember which are logarithms: if the intercept is $c$, then $k={\mathrm{e}}^c$, and if the gradient of the exponential plot is $m$, then $a={\mathrm{e}}^m$. The gradient of the power-law plot is $n$ itself.

Base 10 works in the same way ($\lg y=\lg k+n\lg x$, and then $k={10}^c$); use whichever base the question uses, and do not mix them.

The gradient comes from two points on the line, $m=\frac{Y_2-Y_1}{X_2-X_1}$, and the intercept from $c=Y_1-m{X}_1$ when the line does not reach the axis on the diagram.

Horizontal axis $\ln x$, vertical axis $\ln y$. The gradient is $n$ and the intercept, 2.1, is $\ln k$.