Introduction to Logarithms: Practice

Peel the outer logarithm first by switching it to exponential form. It says the argument $\log_3 x$ equals $4^1$.

$\log_3 x = 4$

Now switch that to exponential form as well.

$x = 3^4 = 81$

Check: $\log_3 81 = 4$ and $\log_4 4 = 1$, as required.

Write the base $9$ as $3^2$ so that the cancellation law can act on the base $3$.

$9^{\log_3 5} = \left(3^2\right)^{\log_3 5} = \left(3^{\log_3 5}\right)^2$

The inner expression collapses, because $3^{\log_3 5} = 5$ by definition of the logarithm.

$\left(3^{\log_3 5}\right)^2 = 5^2 = 25$

Switch to exponential form: the argument must equal $2^2$.

$x^2 - 3x = 4 \quad \Longrightarrow \quad x^2 - 3x - 4 = 0 \quad \Longrightarrow \quad (x - 4)(x + 1) = 0$

So $x = 4$ or $x = -1$. Now test both against the domain rule, which restricts the argument of the logarithm, not $x$ itself.

For $x = 4$ the argument is $16 - 12 = 4 > 0$. For $x = -1$ the argument is $1 + 3 = 4 > 0$. Both are positive, so both values are genuine solutions.

Demand that the argument be positive, then solve the quadratic inequality.

$x^2 - 9 > 0 \quad \Longrightarrow \quad (x - 3)(x + 3) > 0$

The product of the two factors is positive when both are positive ($x > 3$) or both are negative ($x < -3$).

$x < -3 \quad \text{or} \quad x > 3$

Between $-3$ and $3$ the argument is negative, and at $x = \pm 3$ it is zero, so the logarithm exists nowhere in that closed interval.

Set $y = 2^{x-3}$ and solve for $x$. The unknown sits in the exponent, so switch to logarithmic form.

$y = 2^{x-3} \quad \Longrightarrow \quad x - 3 = \log_2 y$

Now isolate $x$ and swap the names of the variables.

$x = \log_2 y + 3 \quad \Longrightarrow \quad f^{-1}(x) = \log_2 x + 3$

Check with a value: $f(5) = 2^2 = 4$, and $f^{-1}(4) = \log_2 4 + 3 = 2 + 3 = 5$.

Let $y = \log_{1/4} 8$, so $\left(\tfrac{1}{4}\right)^y = 8$. Write both sides as powers of $2$, using $\tfrac{1}{4} = 2^{-2}$.

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

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

Trap $100$ between two powers of $5$. Since $5^2 = 25$ and $5^3 = 125$,

$25 < 100 < 125.$

The function $\log_5$ is increasing, so applying it preserves the inequalities.

$\log_5 25 < \log_5 100 < \log_5 125 \quad \Longrightarrow \quad 2 < \log_5 100 < 3$

So the value lies between $2$ and $3$; it is about $2.86$. Ordinary nearness does not tell you where in the interval it lands. The halfway height $2.5$ is reached at $5^{5/2} = 25\sqrt{5} \approx 55.9$, and $100$ sits above that, so this logarithm belongs to the upper half of the interval.

An $x$-intercept has height $0$, so set the expression equal to zero and switch forms.

$\log_2(x - 1) = 0 \quad \Longrightarrow \quad x - 1 = 2^0 = 1$

Solving gives $x = 2$, so the intercept is $(2,\ 0)$.

That is the basic curve's intercept $(1,\ 0)$ shifted one unit right, exactly as the transformation $x \mapsto x - 1$ predicts. The line $x = 1$ is the asymptote, not the intercept.

Compare each with $1$, using the fact that $\log_b b = 1$ and that a logarithm with base bigger than $1$ is increasing.

Since $5 > 2$, we get $\log_2 5 > \log_2 2 = 1$.

Since $2 < 5$, we get $\log_5 2 < \log_5 5 = 1$.

$\log_5 2 < 1 < \log_2 5$

So $\log_2 5$ is the larger. (Trapping confirms it: $2^2 = 4 < 5 < 8 = 2^3$, so $\log_2 5$ is between $2$ and $3$, while $\log_5 2$ is between $0$ and $1$.)

Switch to exponential form: the argument equals $3^4$.

$x^2 = 81 \quad \Longrightarrow \quad x = 9 \ \text{ or } \ x = -9$

Both must be checked against the domain rule, which applies to the argument $x^2$. For $x = 9$ the argument is $81 > 0$, and for $x = -9$ the argument is also $81 > 0$.

Both values are legal, so there are two real solutions. The negative one survives because it is squared before the logarithm ever sees it.

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

$2x - 1 = \log_5 125 = 3$

The last step holds because $5^3 = 125$. Now solve the linear equation.

$2x = 4 \quad \Longrightarrow \quad x = 2$

Check: $5^{2(2) - 1} = 5^3 = 125$.

The point says $\log_4 a = 3$. Switch to exponential form: $a$ is what you get by raising $4$ to the power $3$.

$a = 4^3 = 64$

So the point is $(64,\ 3)$. Note the reflection: $(3,\ 64)$ lies on $y = 4^x$, and swapping the coordinates lands you here.