An exponential function has the form f(x)=a^x

Suppose we want to solve, e.g. 10^x=15. How could we try to solve it?

• Trial and error?;
• Graphical solution?

The logarithmic function lets us calculate 10^x=15 directly, as: 10^x=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 10^x=40 into logarithmic form;

Rewrite log₁₀x=400 in exponential form;

Using the calculator, solve 10^x=3.5

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Logarithms to bases other than 10

Obviously the above logic applies not for just base 10, but for other bases.

In general: y=log_ax <=> a^y=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:

log_aa=1
log_a1=0
log_a(a^x)=x
a^log_ax=x

Worked Examples using any base logarithms

Rewrite 2^x=12 in logarithmic form;

Rewrite log₃x=4 in exponential form and solve;

Simplify

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Important Logarithm Laws

By using the earlier identity and replacing x with a^log_ax (as per previous identity) we can derive the following laws:

• log_a(xy)=log_ax + log_ay;
• log_a(x/y)=log_ax – log_ay;
• log_axⁿ=nlog_ax.

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.

Worked Examples:

• Write as a single logarithm:
  • log₃6 + log₃7;
  • log₂15 – log₂3;
  • 2log₅3 + 3log₅2;
  • log₁₀3 – 4log₁₀(1/2).
• Write in terms of log_ax, log_ay and log_az:
  • 
  • 
  • 
  • 

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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).

Worked Examples

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 4log_x2 – log_x4 = 2.

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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.

Worked Examples

Solve the following equations, giving your answers to 3 decimal places:

• 3^x=20
• 5^4x-1=61
• 5^2x – 12(5^x) + 20 = 0
• 3^x = 2^x+1

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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).

Worked Examples

Solve the following inequalities, giving your answers correct to 3 significant figures: d

• 3^x > 8;
• 5^x < 10;
• 7^2x+5 ≤ 24;
• 0.5^x < 0.001;
• 0.4^x < 0.0004;
• 0.2^x > 25;
• 4^x x 4^3-2x ≤ 1024
• 0.8^2x-5 ≥ 4
• 7^2x+5 ≤ 24
• 0.2^x > 25
• 0.8^1-3x ≥ 10

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Natural Logarithms

In calculus, we will make extensive use of the natural exponential function y=e^x, because its derivative is the same function.

Logarithms with base e are called natural logarithms.

y=e^x <=> x=log_ey. (e is a constant, about 2.718).

We write lny instead of log_ey.

Solve these equations, giving your answers in exact form:

• e^x = 5;
• lnx = 3;
• e^2x+3=7;
• 2lnx+1=5;
• e^2x + 5e^x = 14.

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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=ae^bx, then also lny=lna+bx, and so plotting x against lny gives a line with gradient b and y-intercept lna.

Worked Example

Exercise (N.B. Question 6 should say “the graph of y against x”

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Miscellaneous Questions

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