---
title: Advanced Algebra Practice
---

# Advanced Algebra Practice: Log Rules and Inverses

## Logarithms

### Definitions

First and foremost, we recall our **spiral rules** for converting exponents to logs and logs to exponents:

(D1) Log $\rightarrow$ Exponent
$$
\log_a(b) = c \iff a^c = b
$$
(D2) Exponent $\rightarrow$ Log
$$
a^c = b \iff log_a(b) = c
$$

And we remember the two "special" logs that they'll use to trick you:

(D3) $\log$ without a base means $\log$ **base 10**:
$$
\log(x) = \log_{10}(x)
$$

(D4) $\ln$ means log **base e**[^ln]:
$$
\ln(x) = \log_{e}(x)
$$

### Identities

The above are all definitions[^defns], so now we move to our boxed **identities**! (Mental cheat sheet incoming)

---

1. (Multiplication $\rightarrow$ Addition)
$$
\log( m \cdot n ) = \log(m) + \log(n)
$$
2. (Division $\rightarrow$ Subtraction)
$$
\log \left( \frac{m}{n} \right) = \log(m) - \log(n)
$$
3. (Exponent $\rightarrow$ Multiplication)
$$
\log\left(m^n\right) = n \cdot \log(m)
$$
---

Then there are the "cancellation rules", which we could derive from the three rules above, but for the sake of time I'm just dropping in here as well:

4.
$$
a^{\log_a(x)} = x
$$
5.
$$
\log_a(a^x) = x
$$

You can test your understanding of these last two (along with our definitions) by making sure the following make sense in your head:

$$
e^{\ln(x)} = x
$$
$$
\log(10^x) = x
$$

Finally, we have the "flip rule", which is just helpful for when you need to turn a **fraction** into a single **number**, or vice-versa:

6.
$$
x^{-1} = \frac{1}{x}
$$

7.
$$
\left(\frac{1}{x}\right)^{-1} = x
$$

### Solution Hints

When solving problems using these logarithms, one "trick" you can remember to make your life easier is:
* If the **bases** on both sides of an equation (an equation with **exponents**) are equal, you can **throw away** these bases and just solve for the exponents. For example:

$$
13^{x^2 + 3x} = 13^{\sqrt{3} + x^2}
$$

Looks very scary, but we can make our lives slightly easier by recognizing that both sides are just some exponent with base 13, so that we can throw away the 13s and just solve the equation with the exponents:

$$
x^2 + 3x = \sqrt{3} + x^2
$$

It's still kind of gross, but hopefully at least less scary-looking than the original form, and we can now solve by isolating $x$:

$$
x^2 + 3x = \sqrt{3} + x^2 \implies 3x = \sqrt{3} \implies x = \frac{\sqrt{3}}{3}.
$$

We can do the same thing when there are equations with $\log$ on both sides: if they have the **same base**, we can throw away the logs and just solve the equation based on the stuff **inside** the logs:

$$
\log_{13}(5x+11) = \log_{13}(2 \pi)
$$

turns into

$$
5x + 11 = 2\pi,
$$

allowing us to solve by isolating $x$:

$$
5x + 11 = 2 \pi \implies 5x = 2 \pi - 11 \implies x = \frac{2\pi - 11}{5}.
$$

### Practice Problems

1. Solve for $x$.
$$
4^x = 16
$$
:::spoiler {state="open"} **Click to show/hide solution**
Here we see that, we want to isolate $x$, but it's "trapped" up in the exponent above $4$. But, we see that we can use our "spiral rules" (in this case, definition (D2) from above) to free it out of the exponent:

$$
4^x = 16 \iff \log_4(16) = x
$$

Technically we could leave it like this, since we've now isolated $x$, but we can further simplify, since $\log_4(16)$ is really asking "4 to the power of what equals 16?". We could use our exponent tables, or just start multiplying:

$$
4^0 = 1 \\
4^1 = 4 \\
4^2 = 4 \cdot 4 = 16,
$$

and we've already figured it out: $\log_4(16) = 2$, since $4^2 = 16$. Going back to our original equation where we had isolated $x$, then, we get the final answer:

$$
x = \log_4(16) = 2.
$$

:::

2. Solve for $x$ again.

$$
\left(\frac{1}{64}\right)^{x-2} = 16^{-4x}
$$

:::spoiler {state="open"} **Click to show/hide solution**
Here we don't freak out, even though the problem looks kinda gross, with fractions and negatives and exponents all over the place. Because, we see that the *base* on the left side (1/64) and the *base* on the right side (16) are related to one another. This is promising, since if we can turn this into an equation where both sides have the **same base**, we can just throw out the base and do the math with the exponents equal!

So, let's see how to make the bases the same. The 16 on the right side feels less scary then the left side, so I'll start with that. 16 can be broken down into the square of a simpler number, $4$, so we rewrite the right side as

$$
16^{-4x} = \left(4^2\right)^{-4x},
$$

and since an exponent to another exponent means you multiply the exponents, this gives us

$$
\left(4^2\right)^{-4x} = 4^{-8x}.
$$

Now let's look at the left side. Using the "flip rule", we can turn the fraction into a number, to make our lives slightly easier (since fractions truly hurt my brain):

$$
\left(\frac{1}{64}\right)^{x-2} = \left(64^{-1}\right)^{x-2},
$$

and we again multiply the exponents to make this simpler:

$$
\left(64^{-1}\right)^{x-2} = 64^{(-1)(x-2)} = 64^{-x+2}
$$

Now we're close to both sides having the same base, but we have $64$ as the base on the left but $4$ as the base on the right. 4 is pretty easy to work with, so we probably don't need to break that down any further, but the 64 looks like it can be broken down more, and indeed we find $64$ on our exponent table and see that it's equal to $4^3$! So, the left side term now turns into

$$
64^{-x+2} = \left(4^3\right)^{-x+2} = 4^{(3)(-x+2)} = 4^{-3x + 6}.
$$

So, finally, we have a base of $4$ on either side! Bringing our work together, that is, we have

$$
4^{-3x+6} = 4^{-8x},
$$

so we can throw away the $4$s and just solve the equation with the exponents:

$$
-3x + 6 = -8x \implies 6 = -5x \implies x = -\frac{6}{5}
$$
:::

3. **[Extraneous solutions trickery]** Solve for $x$, check for extraneous solutions.

$$
\log(x+2) + \log(x−1) = 1
$$

:::spoiler {state="open"} **Click to show/hide solution**

Here getting the *potential* solutions is mostly easy: we use our Multiplication $\rightarrow$ Addition identity from above (but in reverse) to turn this *addition* of logs into a *multiplication* inside the log:

$$
\log(x+2) + \log(x-1) = 1 \implies \log((x+2)(x-1)) = 1,
$$

and we FOIL to multiply the two terms inside the log:

$$
\log((x+2)(x-1)) = 1 \implies \log(x^2 + x - 2) = 1
$$

Now we remember that a log written without a base *means* $\log_{10}$, so we can use our "spiral rule" to get rid of the log here:

$$
\log(x^2 + x - 2) = 1 \implies 10^1 = x^2 + x - 2,
$$

so we can obtain an equation (in = 0 form) by subtracting the $10$:

$$
x^2 + x - 12 = 0.
$$

Then we can factor this polynomial, using whatever method we like, and we'll get

$$
(x+4)(x-3) = 0,
$$

which gives us the (**potential**) solutions $x = -4$, $x= 3$. **But**, now we have to go back and check whether or not these solutions are **extraneous**. Since we know that the log of a negative number is *undefined*, we don't really have to plug in and check by solving the whole equation: we can just plug in and see if we obtain any negative numbers inside of logs. So, plugging in our first solution $x = -4$, we get

$$
\log((-4)+2) + \log((-4)-1) = 1 \implies \log(-2) + \log(-5) = 1.
$$

We immediately see the issue: we *thought* that $x=-4$ was one of our solutions, but in fact it produces a negative number inside of *both* logs in our original equation, so it can't be a solution, it turns out: it's **extraneous**.

Next we can check $x=3$:

$$
\log(3+2) + \log(3-1) = 1 \implies \log(5) + \log(2) = 1.
$$

Since this solution doesn't give us any negative numbers inside of logs, we know that it is **not** extraneous: $x = 3$ actually does work as a solution to the original equation, since it doesn't result in any undefined things. So, **after checking for extraneous solutions**, we have the single, unique solution: $x = 3$.

:::

A few more without written-out explanations like the above (but still with solutions!)

4. Solve for $x$.

$$
\log_x(4) = \frac{1}{2}
$$

:::spoiler **Click to show/hide solution**

$$
x = 16
$$

:::

5. Solve for $x$. Check for extraneous solutions.

$$
\log(x-1) + \log(5x) = 2
$$

:::spoiler **Click to show/hide solution**

Original solutions:

$$
x = 5\text{ and }x = -4
$$

**But**, $x = -4$ is an extraneous solution. So the final answer is that there is only one non-extraneous solution:

$$
x = 5.
$$
:::

6. Evaluate.
$$
\log_5(5)
$$

:::spoiler **Click to show/hide solution**

$$
1
$$

:::

7. Evaluate.
$$
\log_{25}(5)
$$

:::spoiler **Click to show/hide solution**

$$
\frac{1}{2}
$$

:::

## Inverse Functions

If you look above, at identities (4) and (5), you'll notice something special about what happens when logs and exponents are combined: they cancel out---*both* the log *and* the exponent go away---so that only $x$ remains. When this happens (with *any* two functions, in fact, not just log and exponents), we call the two functions **inverses** (in our last session I mentioned how it's kind of like protons vs. electrons, how they smash into each other and cancel each other out).

To see an example of inverse functions that don't involve logs or exponents, think about

$$
f(x) = x + 5 \\
g(x) = x - 5
$$

What happens if we *combine* these functions? Meaning, what happens if we plug $g(x)$ into $f(x)$, or vice-versa?

$$
f(g(x)) = f(x - 5) = (x - 5) + 5 = x.
$$

The one function (that subtracts 5 from $x$) cancels out the thing that the other function does (adding 5 to $x$), so that $f(g(x)) = x$. What if we try the other way, plugging $f(x)$ into $g$?

$$
g(f(x)) = g(x + 5) = (x + 5) - 5 = x.
$$

Same thing. That's how we know these two functions are **inverses** of one another.

So, when we're solving an inverse function problem, they'll give you some function that looks like

$$
y = \frac{\sqrt{x}}{2},
$$

and your job is to
1. **Switch $x$ and $y$**, then
2. **Re-solve** this new, switched equation for $y$.

In this case, then, step 1 looks like

$$
x = \frac{\sqrt{y}}{2},
$$

and then step 2 involves isolating $y$ again:

$$
x = \frac{\sqrt{y}}{2} \implies 2x = \sqrt{y} \implies y = (2x)^2 \implies y = 4x^2.
$$

Remember, for that last step, that when there's an exponent at the end of parentheses, and the stuff inside the parentheses is all multiplied together, that means the exponent goes on **each** term inside the parentheses. For example,

$$
(3ab\sqrt{5})^2 = 3^2 a^2 b^2 \sqrt{5}^2 = 9a^2b^25 = 45a^2b^2.
$$

### Practice Problems

1. Find the inverse function.
$$
y = x^3 - 2
$$

:::spoiler {state="open"} **Click to show/hide solution**

Switching $x$ and $y$, we get

$$
x = y^3 - 2,
$$

and then re-solve for $y$:

$$
x = y^3 - 2 \implies y^3 = x + 2 \implies y = \sqrt[3]{x+2}.
$$

Since *cube* roots (unlike square roots) are defined for all numbers, we don't have to worry about any weird extraneous solution stuff here.

:::

2. Find the inverse function, and specify **where it is and isn't defined**.
$$
y = x^2 - 2.
$$

:::spoiler {state="open"} **Click to show/hide solution**

Here we start the same way as the last problem, switching $x$ and $y$ and then solving:

$$
x = y^2 - 2 \implies y^2 = x + 2 \implies y = \sqrt{x+2}.
$$

**But**, now we have to worry about where it's defined and not defined, since square roots are **not** defined for negative numbers. So, looking at the stuff inside the square root here, we see that any $x$ **lower than -2** is going to produce a negative number inside the square root. So, our final answer is the inverse function plus the domain on which it is defined:

$$
y = \sqrt{x+2},\text{ where } x > -2.
$$

:::

A few more, without full explanations but with solutions:

3. Find the inverse.
$$
y = \frac{x}{3} + 2
$$

:::spoiler **Click to show/hide solution**

$$
x = \frac{y}{3} + 2 \implies \frac{y}{3} = x - 2 \implies y = 3(x-2) \implies y = 3x - 6.
$$

:::

4. Find the inverse.
$$
y = (x-1)^3
$$

:::spoiler **Click to show/hide solution**

$$
x = (y-1)^3 \implies y - 1 = \sqrt[3]{x} \implies y = \sqrt[3]{x} + 1.
$$

:::

[^ln]: If it helps, you can remember that $\ln$ stands for "**L**ogarithmus **N**aturali", the Latin for "Natural Logarithm", since it was created back in Latin times.

[^defns]: *Definitions* are like, what do they *mean* when they say $\ln$, whereas *identities* are like, how you actually solve the different problems. You need both to solve the problems, but step-by-step you use definitions first, then identities. If you see a problem like "$\ln(3x) = 5$, solve for $x$", you first use the definition to see that $\ln(3x)$ *means* $\log_e(3x)$, then use the identities to solve this: $\ln(3x) = 5 \iff e^5 = 3x \iff x = \frac{e^5}{3}$.