Introduction to Logarithms: Practice

The definition of the logarithm is a two-way street. The statement $\log_b x = y$ says that $y$ is the exponent you put on the base $b$ to get $x$, so it is the same as $b^y = x$.

Here the base is $2$, the result is $32$, and the exponent is $5$.

$2^5 = 32$

The base stays the base and the value of the logarithm is the exponent, so $2^5 = 32$ is the matching exponential form.

Ask the defining question: three to the what gives eighty-one?

Run up the powers of $3$: $3^1 = 3$, $3^2 = 9$, $3^3 = 27$, $3^4 = 81$.

$3^4 = 81 \quad \Longrightarrow \quad \log_3 81 = 4$

The answer to a logarithm is the exponent, never the number you started with.

Ask: five to the what gives twenty-five?

$5^2 = 25 \quad \Longrightarrow \quad \log_5 25 = 2$

So the logarithm is $2$, the exponent that turns the base $5$ into $25$.

Ask: ten to the what gives one thousand? Counting the zeros answers it.

$10^3 = 1000 \quad \Longrightarrow \quad \log_{10} 1000 = 3$

For base $10$ the logarithm of a power of ten is just the number of zeros.

Ask the defining question: $b$ to the what gives $1$? Every nonzero number raised to the power $0$ equals $1$.

$b^0 = 1 \quad \Longrightarrow \quad \log_b 1 = 0$

This is why every logarithm graph crosses the $x$-axis at the point $(1,\ 0)$, whatever the base.

Ask: nine to the what gives nine? Raising a number to the first power leaves it unchanged.

$9^1 = 9 \quad \Longrightarrow \quad \log_9 9 = 1$

In general $\log_b b = 1$ for every legal base.

Switch to exponential form. The statement says the exponent that turns $4$ into $x$ is $3$.

$x = 4^3 = 64$

Check it: $\log_4 64$ asks for the exponent taking $4$ to $64$, and $4^3 = 64$, so the answer is $3$ as required.

The inverse of an exponential undoes it: it receives a positive number and returns the exponent that produced it. That is exactly what the logarithm to the same base does.

$f(x) = 3^x \quad \Longrightarrow \quad f^{-1}(x) = \log_3 x$

The base of the exponential becomes the base of the logarithm. Notice that $x^3$ is a power function, not an exponential, so reversing it is a cube root, not a logarithm.

Use the cancellation law. Asking for $\log_6(6^4)$ means asking for the exponent you put on $6$ to get $6^4$, and the exponent $4$ obviously does the job. Since $6^x$ is one-to-one, no other exponent does.

$\log_6\!\left(6^4\right) = 4$

In general $\log_b(b^t) = t$ for every real $t$, because the logarithm and the exponential undo each other.

Read the exponent for what it is. By definition, $\log_5 11$ is the exponent you must put on $5$ to get $11$.

So putting it on $5$ gives $11$.

$5^{\log_5 11} = 11$

This is the other cancellation law, $b^{\log_b x} = x$, valid for every $x > 0$.

Crossing the $x$-axis means the height is zero, so solve $\log_b x = 0$. Switching to exponential form,

$x = b^0 = 1$

So the crossing happens at $(1,\ 0)$, and it happens there for every base, because $b^0 = 1$ no matter what $b$ is. The point $(b,\ 1)$ does lie on the graph, but its height is $1$, not $0$, so it is not the $x$-intercept.

Ask: two to the what gives sixty-four? Run up the powers of $2$: $2, 4, 8, 16, 32, 64$.

$2^6 = 64 \quad \Longrightarrow \quad \log_2 64 = 6$

The sixth power of $2$ is $64$, so the logarithm is $6$.