Logarithms: Find the Missing Power

Which power lands on 10?

Start with a familiar calculation: $2\times2\times2=8$. In $2^3=8$, the base is 2, the exponent is 3 and the result is 8. Now cover the exponent and ask: “2 raised to which power gives 8?” The answer is still 3.

A logarithm answers precisely that question. We write $\log_2 8=3$, read as “log base 2 of 8 is 3”. The logarithm’s answer is an exponent, not the result of multiplying 2 by 8.

Integer powers do not fill every gap. Since $2^3=8$ and $2^4=16$, the power that gives 10 lies between 3 and 4. We name it $\log_2 10$. That is an exact value; a decimal such as 3.3219 is an approximation.

In the explorer, predict the interval first. Then move the target and watch the missing exponent respond. For base 2, a larger positive target needs a larger exponent. The target need not be a whole number.

Read the same fact in two directions

The exponential form and the logarithmic form describe one fact:

The base stays the same. The result of the power, $x$, becomes the argument of the logarithm—the number inside it. The exponent, $y$, becomes the answer. For real logarithms, we use $a>0$, $a\ne1$ and $x>0$; the next topic explains why.

Track the roles rather than the letters. If the statement is $b^p=q$, its equivalent is $\log_b q=p$. Changing a letter does not change the relationship.

These operations can also undo one another:

Both rules need a valid base. They are inverse operations, not permission to cross out matching symbols anywhere. The logarithm must be the exponent in the first expression; in the second, its whole argument must be the matching power.

Three checks before you calculate

A real logarithm $\log_a x$ needs a positive base, a base other than 1, and a positive argument:

Why exclude 1? Every real power of 1 is 1. The equation $1^t=7$ has no answer, while $1^t=1$ has every real number as an answer. Neither gives a unique missing exponent. Thus $\log_1 1$ is undefined too.

Zero is not a suitable base: its positive powers remain zero, and its negative powers are undefined. Negative bases have some familiar integer powers, such as $(-2)^3=-8$, but do not give a real exponential function for every real exponent; $(-2)^{1/2}$, for example, is not real.

For an allowed positive base, every real power is positive. That is why the argument cannot be zero or negative. The answer of a logarithm can still be negative or zero.

Check the whole argument. In $\log_2(x+4)$, the requirement is $x+4>0$, so $x>-4$. It is not the stronger and unnecessary restriction $x>0$.

A negative answer can be perfectly valid

“Positive and not 1” includes every base between 0 and 1. The allowed bases form $(0,1)\cup(1,\infty)$. Replacing this with $a>1$ would throw away an entire family of logarithms.

Use negative exponents carefully. For a nonzero positive base, $a^{-n}=1/a^n$. Thus $2^{-3}=1/8$, not $-8$. The exponent is negative; the result is still positive. Consequently $\log_2(1/8)=-3$ is a valid logarithm.

A base below 1 reverses the pattern. Its positive powers shrink: $(1/2)^1=1/2$, $(1/2)^2=1/4$. Its negative powers grow: $(1/2)^{-1}=2$, $(1/2)^{-2}=4$.

Two reliable anchors work for every valid base: $\log_a1=0$, because $a^0=1$, and $\log_a a=1$, because $a^1=a$. Between these anchors and familiar powers, you can often predict a logarithm’s sign before calculating it. Distinguish three different things each time: the base, the argument and the answer.

Two bases you will meet often: 10 and e

A logarithm always has a base, even when notation hides it. The common logarithm uses base 10: $\log_{10}x$. In this course, an unqualified $\log x$ means $\log_{10}x$. When using another book or a calculator, check its convention.

The natural logarithm uses the number $e$, approximately 2.71828. Its notation is $\ln x=\log_e x$, read “natural log of x”. It answers the same missing-exponent question: $\ln x=y$ means $e^y=x$. Natural logs will become especially useful on the Class 12 calculus route.

Do not mix the two buttons or treat $e$ as a variable to choose. For example, $\log100=2$, while $\ln(e^2)=2$. Their answers happen to match because the arguments are matching powers of their respective bases.

Keep exact expressions when possible. Write $\ln2$ as the exact answer and use $\ln2\approx0.6931$ only when an approximation is useful. Both common and natural logs require positive arguments. Their input 1 gives zero, and an input between 0 and 1 gives a negative output.