An exponential function has the form f(x)=ax
Suppose we want to solve, e.g. 10x=15. How could we try to solve it?
- Trial and error?;
- Graphical solution?
The logarithmic function lets us calculate 10x=15 directly, as: 10x=15 <=> x=log1015.
We can read this as saying that x is the number that 10 must be raised to to get 15.
We call 10 the base and x the index.
We can find a numerical solution using the calculator button: [ log[…][…] ]
Worked Examples Using Base 10 Logarithms
Rewrite 10x=40 into logarithmic form;
Rewrite log10x=400 in exponential form;
Using the calculator, solve 10x=3.5
Logarithms to bases other than 10
Obviously the above logic applies not for just base 10, but for other bases.
In general: y=logax <=> ay=x. This is defined provided that a>0, a is not equal to 1 and x>0 (why are these conditions necessary?)
Some obvious identities that must nevertheless be memorised are as follows:
Worked Examples using any base logarithms
Rewrite 2x=12 in logarithmic form;
Rewrite log3x=4 in exponential form and solve;
Important Logarithm Laws
By using the earlier identity and replacing x with alogax (as per previous identity) we can derive the following laws:
- loga(xy)=logax + logay;
- loga(x/y)=logax – logay;
These are very important laws that will be used extensively across P3 topics and so should be memorised immediately. As well as understanding the derivation, you should immediately recognise when one of these laws can be applied.
- Write as a single logarithm:
- log36 + log37;
- log215 – log23;
- 2log53 + 3log52;
- log103 – 4log10(1/2).
- Write in terms of logax, logay and logaz:
Solving logarithmic equations
We’ve already been doing this, but now we can start to make them more complicated by relying on our logarithm laws.
If possible we should try to get all terms in the same base, so for instance if we have log27+5=log2x, we can change the 5 into log232 to make things easier.
We should test solutions to the original equations to ensure they satisfy the conditions (i.e. x>0, a>0, a not 1).
1.) Solve the equation log104 + 2log10x = 2;
2.) Solve the equation log3(x+11) – log3(x-5) = 2;
3.) Solve 2log8(x+2) = log8(2x + 19);
4.) Solve 4logx2 – logx4 = 2.
Using logarithms to solve exponential equations with different bases
The two main “new” methods we will use are:
- Taking the logarithm of each side of the equation and then using algebraic manipulation to isolate the variable; and
- Substituting and then solving related quadratic equations.
Solve the following equations, giving your answers to 3 decimal places:
- 52x – 12(5x) + 20 = 0
- 3x = 2x+1
Solving Exponential Inequalities
As with solving exponential equations we take the logarithm of both sides of the equation (to ensure the variable is not in the index).
However, if the base is less than 1, the inequality sign will be reversed (because the log function in this case is a decreasing function).
Also, remember that dividing or multiplying by a negative reverses the inequality sign (any argument less than 1 gives a negative because log1=0).
Solve the following inequalities, giving your answers correct to 3 significant figures: d
- 3x > 8;
- 5x < 10;
- 72x+5 ≤ 24;
- 0.5x < 0.001;
- 0.4x < 0.0004;
- 0.2x > 25;
- 4x x 43-2x ≤ 1024
- 0.82x-5 ≥ 4
- 72x+5 ≤ 24
- 0.2x > 25
- 0.81-3x ≥ 10
In calculus, we will make extensive use of the natural exponential function y=ex, because its derivative is the same function.
Logarithms with base e are called natural logarithms.
y=ex <=> x=logey. (e is a constant, about 2.718).
We write lny instead of logey.
Solve these equations, giving your answers in exact form:
- ex = 5;
- lnx = 3;
- e2x + 5ex = 14.
Using Logarithms to Linearise Data
Data sets that form a curve can often be transformed in order to represent them with a line by plotting x or lnx against y.
e.g. If data is modelled by y=aebx, then also lny=lna+bx, and so plotting x against lny gives a line with gradient b and y-intercept lna.
Exercise (N.B. Question 6 should say “the graph of y against x”