1A, 1B - Logarithms as exponents

Reversing exponentiation

A logarithm tells you the exponent when the base and the result of a power are known.

Since $2^3=8$, the power of $2$ needed to make $8$ is $3$. We write this as $\log_2 8=3$, read “logarithm of eight to base two equals three”. The small $2$ is the base, $8$ is the argument, and the value of the logarithm is the exponent $3$.

The two statements contain exactly the same information. Keep the base fixed when changing form: the logarithm's value becomes the exponent, and its argument becomes the result of the power.

Now reverse the direction and use a negative exponent. A negative logarithm is possible even though a negative argument is not.

Every real power of a positive base is positive. Therefore no real exponent of $5$ produces $0$ or $-25$: $\log_5 0$ and $\log_5(-25)$ have no real values. The restriction is on the argument, not on the logarithm's answer.

Exact logarithms

To evaluate a logarithm exactly, ask which exponent gives its argument.

The zero and first powers immediately give two useful values.

The logarithm of $1$ is therefore $0$, not $1$. A logarithm whose argument equals its base has value $1$.

Keep the base $4$, but change the argument to a number between $0$ and $1$. Fractional and negative indices now work together.

For any base greater than $1$, arguments above $1$ have positive logarithms, and arguments between $0$ and $1$ have negative logarithms. For example, $2^2<6<2^3$ tells us $2<\log_2 6<3$, even without its decimal value. A logarithm need not be an integer or a simple fraction.

Logarithmic graphs

Reversing exponentiation swaps the input and output of an exponential graph.

The point $(3,8)$ lies on $y=2^x$ because $2^3=8$. The equivalent statement $\log_2 8=3$ puts $(8,3)$ on $y=\log_2 x$. Swapping every pair of coordinates reflects the graph in the line $y=x$.

[DIAGRAM: asset_name: Exponential and logarithmic graphs are reflections; asset_slug: logarithm-reflection; caption: Swapping coordinates reflects an exponential graph in the line y equals x to give its logarithmic graph; description: Left panel uses equal coordinate scales to show y equals 2 to the x in blue, y equals log base 2 of x in sage and the dashed reflection line y equals x. The points (1,2) and (2,1) are connected across the reflection line, and the intercepts (0,1) and (1,0) are marked. Right panel shows y equals log base 3 of x with points (1/3,-1), (1,0), (3,1) and (9,2), approaching the vertical asymptote x equals 0 from the right.]

For $y=\log_b x$, with natural-number $b>1$:

• the domain is $x>0$, because the exponential's outputs are positive;
• the range is all real numbers, because the exponential accepts every real exponent;
• the curve crosses the horizontal axis at $(1,0)$ and passes through $(b,1)$;
• it has no vertical-axis intercept and has vertical asymptote $x=0$;
• it increases throughout its domain, becoming less steep as $x$ increases.

As $x$ approaches $0$ from the right, the logarithm decreases without bound. As $x$ increases without bound, the logarithm increases without bound, however slowly. These are shape statements; no gradient formula is needed.

Now change the base to $3$. The standard shape and intercept remain, but the point with output $1$ moves.

Use exact points to position the curve, then draw its smooth shape; joining the points with straight segments would misrepresent the function.

Quick Recap

The opening answers are $\log_2 8=3$ and $\log_2(1/8)=-3$, because $2^3=8$ and $2^{-3}=1/8$. A logarithm gives an exponent, accepts a positive argument and can have a negative value. Its graph swaps the coordinates of the exponential graph and has vertical asymptote $x=0$.