Introduction to Logarithms: Additional Practice (Set 2)

Work from the inside out. The inner logarithm asks: two to the what gives eight?

$2^3 = 8 \quad \Longrightarrow \quad \log_2 8 = 3$

Now the expression is $\log_3 3$, and a base raised to the first power is itself.

$\log_3 3 = 1$

So the value is $1$.

Let $y = \log_{16} 32$, so $16^y = 32$. Both sides are powers of $2$, so rewrite them.

$2^{4y} = 2^5$

Two powers of $2$ are equal only when their exponents agree, because $2^x$ is one-to-one.

$4y = 5 \quad \Longrightarrow \quad y = \frac{5}{4}$

As a check, $16 < 32 < 16^2 = 256$, and $\log_{16}$ is increasing, so the value has to lie strictly between $1$ and $2$. That rules out $2$ itself as well as anything below $1$, leaving $\tfrac{5}{4}$.

Let $y = \log_{1/2} 8$, so $\left(\tfrac{1}{2}\right)^y = 8$. A base less than $1$ must be raised to a negative power to give a result greater than $1$.

$\left(\frac{1}{2}\right)^{-3} = 2^3 = 8$

So $\log_{1/2} 8 = -3$. This matches the graph: with $0 < b < 1$ the logarithm is decreasing, so arguments bigger than $1$ have negative logarithms.

Trap $500$ between two powers of $3$. The powers run $3, 9, 27, 81, 243, 729$, so

$243 < 500 < 729 \quad \text{that is} \quad 3^5 < 500 < 3^6.$

The function $\log_3$ is increasing, so applying it preserves the inequalities.

$5 < \log_3 500 < 6$

So the value lies between $5$ and $6$.

Switch to exponential form: the argument must equal $3^1$.

$x^2 + 2x = 3 \quad \Longrightarrow \quad x^2 + 2x - 3 = 0 \quad \Longrightarrow \quad (x + 3)(x - 1) = 0$

So $x = -3$ or $x = 1$. Check the argument in each case, since it is the argument that must be positive.

For $x = 1$ the argument is $1 + 2 = 3 > 0$. For $x = -3$ the argument is $9 - 6 = 3 > 0$. Both give $\log_3 3 = 1$, so both are solutions.

Set $y = 3^x + 4$ and unwind the operations in reverse. First undo the addition.

$y - 4 = 3^x$

Now the unknown is in the exponent, so switch to logarithmic form.

$x = \log_3(y - 4) \quad \Longrightarrow \quad f^{-1}(x) = \log_3(x - 4)$

Check with a value: $f(2) = 9 + 4 = 13$, and $f^{-1}(13) = \log_3 9 = 2$. Notice the domain of the inverse is $x > 4$, which is exactly the range of $f$.

The logarithm is the inverse of $7^x$, and the range of an inverse function is the domain of the original.

The domain of $7^x$ is every real number, since you may raise $7$ to any real power.

$\text{range of } \log_7 x = \text{domain of } 7^x = \text{all real numbers}$

So a logarithm can be negative, zero, or positive: it is the argument that is restricted, never the value.

Switch to exponential form: the argument equals $2^3$.

$x^2 + 1 = 8 \quad \Longrightarrow \quad x^2 = 7 \quad \Longrightarrow \quad x = \pm\sqrt{7}$

Check the argument for each root. It is $8$ in both cases, which is positive, so the logarithm is defined and both roots are genuine.

There are two real solutions, $x = \sqrt{7}$ and $x = -\sqrt{7}$.

The graph of $y = b^x$ has the horizontal asymptote $y = 0$, and reflecting across the line $y = x$ turns a horizontal line into a vertical one.

$y = 0 \ \text{ (asymptote of } b^x) \ \longrightarrow \ x = 0 \ \text{ (asymptote of } \log_b x)$

So the logarithm dives down beside the $y$-axis without ever touching it, which is exactly why it is undefined at $x = 0$ and why its domain is only $x > 0$. It has no horizontal asymptote at all: it keeps climbing, however slowly, without bound.

Let $y = \log_4 \tfrac{1}{32}$, so $4^y = \tfrac{1}{32}$. Write both sides as powers of $2$, using $\tfrac{1}{32} = 2^{-5}$.

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

Check: $4^{-5/2} = \dfrac{1}{4^{5/2}} = \dfrac{1}{\left(4^{1/2}\right)^5} = \dfrac{1}{32}$.

Switch to exponential form. The statement says $x$ raised to the fourth power gives $16$.

$x^4 = 16$

The real roots are $x = 2$ and $x = -2$, since $(-2)^4 = 16$ as well. But a logarithm base must be positive and not equal to $1$, so the negative root is not admissible.

The only solution is $x = 2$, and it checks out: $\log_2 16 = 4$ because $2^4 = 16$.

Evaluate the three easy ones first. Since $3^2 = 9$, we have $\log_3 9 = 2$. Since $9^1 = 9$, we have $\log_9 9 = 1$. Since $9^{1/2} = 3$, we have $\log_9 3 = \tfrac{1}{2}$.

Now trap the remaining one between powers of $2$, using $2^3 = 8$ and $2^4 = 16$.

$8 < 9 < 16 \quad \Longrightarrow \quad 3 < \log_2 9 < 4$

So $\log_2 9$ is more than $3$, which beats $2$, $1$, and $\tfrac{1}{2}$. For an argument greater than $1$, a smaller base must be raised to a bigger power to reach it, so the smaller base produces the larger logarithm. That rule needs the argument to exceed $1$: for an argument between $0$ and $1$ the logarithms are negative and the comparison reverses.