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Laws of Logarithms for University Admission

Updated August 2025

Logarithms are the inverse of exponential functions, allowing us to determine the power to which a base must be raised to produce a given value. This page covers the fundamental logarithmic laws, their graphical representations, and the necessary constraints for TMUA and ESAT preparation.

Core concept

A logarithm is defined by the equivalence ab=cb=logaca^b = c \Leftrightarrow b = \log_a c, representing the power bb needed for base aa to reach value cc.

Logarithms are very closely related to indices. In fact, they are really the inverse of indices: they tell you what power a number has to be raised to, rather than raising a number to a power. Historically, logarithms were invented to make complex calculations easier before the existence of calculators, by converting multiplication into addition.

The Fundamental Definition

The base of a logarithm, written as a subscript, identifies the number being raised to a power. For example, log10\log_{10} tells you what power 10 needs to be raised to in order to get a given number:

  1. log1010=1\log_{10} 10 = 1 because 101=1010^1 = 10.
  2. log10100=2\log_{10} 100 = 2 because 102=10010^2 = 100.
  3. log101000=3\log_{10} 1000 = 3 because 103=100010^3 = 1000.
  4. log10271.43136\log_{10} 27 \approx 1.43136 because 101.43136=2710^{1.43136\dots} = 27.

We can use any base aa. For example, log2\log_2 tells you what power 2 needs to be raised to reach a value:

  1. log232=5\log_2 32 = 5 because 25=322^5 = 32.
  2. log212=1\log_2 \frac{1}{2} = -1 because 21=122^{-1} = \frac{1}{2}.

In general, the relationship is ab=cb=logaca^b = c \Leftrightarrow b = \log_a c. There are three critical restrictions to remember for the TMUA: the base aa must be positive (a>0a > 0) and not equal to 1, the number we take the log of (cc) must be positive (c>0c > 0), but the resulting log (bb) can be any real number, including zero or negative values.

Graphical Representation

We can understand logarithms by looking at them as the inverse of exponential graphs. Consider y=2xy = 2^x. This function takes an input from the xx axis and provides a yy value. If we start on the yy axis and trace back to the xx axis, we are finding the logarithm.

img-47.jpeg

The graph of y=log2xy = \log_2 x is simply the graph of y=2xy = 2^x with the xx and yy axes swapped. This is a reflection in the line y=xy = x.

img-48.jpeg

Note that the log graph is only defined for x>0x > 0 and it always crosses the xx axis at 1, because a0=1a^0 = 1 for any base aa, meaning loga1=0\log_a 1 = 0.

The Laws of Logarithms

You must be fluent in manipulating logarithms using these four primary rules. Their derivations are rooted in the laws of indices.

1. The Product Rule: logax+logay=loga(xy)\log_a x + \log_a y = \log_a(xy) This is the logarithmic equivalent of apaq=ap+qa^p a^q = a^{p+q}. We can see this works because alogax+logay=alogaxalogay=xy=aloga(xy)a^{\log_a x + \log_a y} = a^{\log_a x} a^{\log_a y} = xy = a^{\log_a(xy)}.

2. The Quotient Rule: logaxlogay=loga(xy)\log_a x - \log_a y = \log_a (\frac{x}{y}) This corresponds to the index law apaq=apq\frac{a^p}{a^q} = a^{p-q}. Derivation: alogaxlogay=alogaxalogay=xy=aloga(xy)a^{\log_a x - \log_a y} = \frac{a^{\log_a x}}{a^{\log_a y}} = \frac{x}{y} = a^{\log_a (\frac{x}{y})}.

3. The Power Rule: klogax=loga(xk)k \log_a x = \log_a (x^k) This follows from (ap)k=apk(a^p)^k = a^{pk}. Derivation: aklogax=(alogax)k=xk=aloga(xk)a^{k \log_a x} = (a^{\log_a x})^k = x^k = a^{\log_a (x^k)}.

4. Special Cases:

  • The Reciprocal Rule: loga(1x)=logax\log_a (\frac{1}{x}) = -\log_a x. This is a specific instance of the power rule where k=1k = -1.
  • Base Identity: logaa=1\log_a a = 1, because a1=aa^1 = a.

Change of Base Formula

Although questions specifically requiring the change of base formula will not be set in the TMUA, it is a useful tool for your mathematical kit. It allows you to convert a logarithm from one base to another:

logab=logcblogca\log_a b = \frac{\log_c b}{\log_c a}

A useful result of this is that logab=1logba\log_a b = \frac{1}{\log_b a}. You can derive this by letting p=logabp = \log_a b, which means ap=ba^p = b. Taking logc\log_c of both sides gives logcap=logcb\log_c a^p = \log_c b, or plogca=logcbp \log_c a = \log_c b, leading to p=logcblogcap = \frac{\log_c b}{\log_c a}.

Key takeaways

  • The fundamental identity is ab=cb=logaca^b = c \Leftrightarrow b = \log_a c.
  • Logarithms are only defined for positive arguments (x>0x > 0) and positive bases (a>0,a1a > 0, a \neq 1).
  • Addition of logs corresponds to multiplication of arguments: logax+logay=loga(xy)\log_a x + \log_a y = \log_a (xy).
  • Subtraction of logs corresponds to division of arguments: logaxlogay=loga(xy)\log_a x - \log_a y = \log_a (\frac{x}{y}).
  • The power rule allows coefficients to move to the exponent: klogax=loga(xk)k \log_a x = \log_a (x^k).
Tips

When solving logarithmic equations in the TMUA, always check your final answers against the original constraints. If a solution results in taking the log of a negative number or zero, it must be discarded.

Cautions

The most common mistake is forgetting that the argument of a log must be strictly positive. Always ensure x>0x > 0 in logax\log_a x.

Insight

The logarithm is the only function that transforms multiplication into addition. This property makes it fundamentally important in fields ranging from complexity theory in computer science to the measurement of sound (decibels) and earthquakes (Richter scale).

Worked Examples

Example 1
Find the real non-zero solution to the equation 2(9x)8(3x)=14\frac{2^{(9^x)}}{8^{(3^x)}} = \frac{1}{4}
A:log32\log_3 2
B:2log322\log_3 2
C:1
D:2
E:log23\log_2 3
F:2log232\log_2 3

Practice Questions

Practice Question 1
Which one of the following numbers is largest in value?
(All angles are given in radians.)
A:tan(3π4)\tan \left(\frac{3\pi}{4}\right)
B:log10100\log_{10} 100
C:sin10(π2)\sin^{10} \left(\frac{\pi}{2}\right)
D:log210\log_2 10
E:(21)20(\sqrt{2}-1)^{20}

Frequently asked questions

Can the result of a logarithm be negative?

Yes. While the base aa and the argument xx must be positive, the result bb can be any real number. For example, log20.5=1\log_2 0.5 = -1.

Why is loga1\log_a 1 always zero?

Because any positive base aa raised to the power of 0 equals 1 (a0=1a^0 = 1), the inverse operation must yield loga1=0\log_a 1 = 0.

Is loga(x+y)\log_a(x + y) equal to logax+logay\log_a x + \log_a y?

No. This is a common error. The correct law is loga(xy)=logax+logay\log_a(xy) = \log_a x + \log_a y. There is no simple law for expanding the logarithm of a sum.

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