Introduction to Logarithms: Practice

Let $y = \log_8 4$, so that $8^y = 4$. Both $8$ and $4$ are powers of $2$, so write them that way.

$8^y = 4 \quad \Longrightarrow \quad 2^{3y} = 2^2$

Two powers of $2$ are equal only when their exponents are equal, because $2^x$ is one-to-one. So $3y = 2$ and $y = \tfrac{2}{3}$.

Check: $8^{2/3} = \left(8^{1/3}\right)^2 = 2^2 = 4$.

Let $y = \log_{1/3} 9$, so $\left(\tfrac{1}{3}\right)^y = 9$. A base smaller than $1$ needs a negative exponent to produce a result bigger than $1$.

$\left(\frac{1}{3}\right)^{-2} = 3^2 = 9$

So $\log_{1/3} 9 = -2$. This is the picture of a decreasing logarithm: with $0 < b < 1$, big arguments have negative logarithms.

The domain restriction applies to the argument of the logarithm, which here is $x - 4$. Demand that it be positive.

$x - 4 > 0 \quad \Longrightarrow \quad x > 4$

The endpoint $x = 4$ is excluded because $\log_3 0$ does not exist: no exponent turns $3$ into $0$. The graph has a vertical asymptote at $x = 4$.

Trap $50$ between two powers of $2$. Since $2^5 = 32$ and $2^6 = 64$,

$32 < 50 < 64.$

The function $\log_2$ is increasing, so applying it to all three parts keeps the inequalities pointing the same way.

$\log_2 32 < \log_2 50 < \log_2 64 \quad \Longrightarrow \quad 5 < \log_2 50 < 6$

So $\log_2 50$ lies between $5$ and $6$ (it is about $5.64$).

The graph of an inverse function is the original reflected across the line $y = x$, and reflecting a point across that line swaps its coordinates.

$(2,\ 25) \ \longrightarrow \ (25,\ 2)$

Check it directly against the definition: $5^2 = 25$, so the exponent that turns $5$ into $25$ is $2$, which says $\log_5 25 = 2$.

By definition, $\log_{10} 3.5$ is the exponent you must put on $10$ to get $3.5$. Putting it on $10$ therefore gives back $3.5$.

$10^{\log_{10} 3.5} = 3.5$

This is the cancellation law $b^{\log_b x} = x$, valid for every $x > 0$. You never need to know the exponent's decimal value to use it.

The definition says $\log_b x = y$ means $b^y = x$. The base of the logarithm is the base of the power, and the value of the logarithm is the exponent.

Here $b$ is the base, $3$ is the value, and $40$ is the argument.

$b^3 = 40$

Everything else scrambles the three roles.

The graph of $y = \log_3 x$ has a vertical asymptote at $x = 0$, the reflection of the exponential's horizontal asymptote. The asymptote sits wherever the argument reaches $0$.

$x + 6 = 0 \quad \Longrightarrow \quad x = -6$

So the curve dives down beside the vertical line $x = -6$, and the domain is $x > -6$. An asymptote of a logarithm is a vertical line, so an answer of the form $y = \text{constant}$ cannot be right.

Let $y = \log_{25} 125$, so $25^y = 125$. Write both sides as powers of $5$.

$5^{2y} = 5^3 \quad \Longrightarrow \quad 2y = 3 \quad \Longrightarrow \quad y = \frac{3}{2}$

Check: $25^{3/2} = \left(25^{1/2}\right)^3 = 5^3 = 125$. Since $125$ is bigger than the base $25$, the logarithm must exceed $1$, which $\tfrac{3}{2}$ does.

The domain of $\log_2$ is the set of positive numbers, since those are exactly the outputs of $2^x$.

No real exponent makes $2^y = 0$. The powers of $2$ shrink toward $0$ as $y$ runs to the left, but they stay strictly positive and never land on it.

$2^y > 0 \ \text{ for every real } y \quad \Longrightarrow \quad \log_2 0 \ \text{ does not exist}$

The other three are all positive: $\log_2 \tfrac{1}{64} = -6$, $\log_2 1 = 0$, and $\log_2 1024 = 10$.

The inverse of $f(x) = 2^x$ is $f^{-1}(x) = \log_2 x$, so $f^{-1}(64) = \log_2 64$: the exponent that turns $2$ into $64$.

$2^6 = 64 \quad \Longrightarrow \quad f^{-1}(64) = 6$

You can also read it as undoing $f$: since $f(6) = 64$, the inverse must send $64$ back to $6$.

The unknown is stuck in the exponent, so switch to logarithmic form: $x = \log_3 \tfrac{1}{243}$.

Now ask the defining question. Since $3^5 = 243$, the reciprocal needs the negative exponent.

$3^{-5} = \frac{1}{3^5} = \frac{1}{243}$

So $x = -5$.