Logarithm Proofs Practice- Precalculus Exercises

What You're Actually Learning When You Do Logarithm Proofs

Let's be clear: logarithm proofs aren't busy work. They're training your brain to see how mathematical expressions relate to each other. When you prove that log(ab) = log(a) + log(b), you're not just memorizing a formula—you're understanding why the relationship exists.

Most students rush through these problems to get to the next section. That's a mistake. The people who actually understand precalculus are the ones who can derive relationships, not just apply them.

The Core Logarithm Properties You Need to Prove

Every logarithm proof problem boils down to one or more of these fundamental relationships:

Your job is to demonstrate that these statements are always true using the definition of logarithms and basic algebra.

How to Approach Logarithm Proof Problems

Here's the actual process that works:

Step 1: Write Down What You Know

If you're proving logb(xy) = logb(x) + logb(y), start by defining x and y in terms of the base b.

Let x = bm and y = bn

This is non-negotiable. You're translating the problem into exponential form, which gives you something concrete to work with.

Step 2: Apply the Definition

From x = bm, you get logb(x) = m

From y = bn, you get logb(y) = n

Step 3: Work the Left Side

xy = bm · bn = bm+n

Therefore, logb(xy) = m + n

Step 4: Substitute What You Found

m + n = logb(x) + logb(y)

Done. The proof is complete.

Worked Examples

Example 1: Prove the Quotient Rule

Statement: logb(x/y) = logb(x) − logb(y)

Proof:

Let x = bm and y = bn

Then logb(x) = m and logb(y) = n

x/y = bm / bn = bm−n

logb(x/y) = m − n = logb(x) − logb(y) ✓

Example 2: Prove the Power Rule

Statement: logb(xn) = n · logb(x)

Proof:

Let x = bm, so logb(x) = m

xn = (bm)n = bmn

logb(xn) = mn = n · m = n · logb(x) ✓

Example 3: Prove the Change of Base Formula

Statement: loga(x) = logb(x) / logb(a)

Proof:

Let loga(x) = y, so ay = x

Take logb of both sides:

logb(ay) = logb(x)

y · logb(a) = logb(x)

y = logb(x) / logb(a)

Substitute y back: loga(x) = logb(x) / logb(a) ✓

Practice Problems

Try these before checking the solutions. No peeking early.

Problem 1

Prove: logb(1) = 0

Hint: Remember that b0 = 1 for any base.

Problem 2

Prove: logb(b) = 1

Hint: What exponent turns b into b?

Problem 3

Prove: blogb(x) = x

Hint: This is the definition of logarithms in exponential form.

Problem 4

Prove: logb(a) = 1 / loga(b)

Hint: Start with logb(a) = y and solve for y.

Problem 5

Prove: logb(x) = logb(c) · logc(x)

Hint: Use the change of base formula twice.

Solutions

Solution 1

b0 = 1 for all b ≠ 0

Taking logb of both sides: logb(1) = 0 ✓

Solution 2

b1 = b

Taking logb of both sides: logb(b) = 1 ✓

Solution 3

Let logb(x) = y

By definition: by = x

Substitute y back: blogb(x) = x ✓

Solution 4

Let logb(a) = y

Then by = a

Taking loga of both sides: y · loga(b) = 1

y = 1 / loga(b)

Substitute y back: logb(a) = 1 / loga(b) ✓

Solution 5

Using change of base on logc(x):

logc(x) = logb(x) / logb(c)

Rearrange: logb(x) = logb(c) · logc(x) ✓

Common Mistakes That Cost You Points

These errors show up constantly in precalculus exams:

Quick Reference: Logarithm Properties

PropertyFormulaKey Strategy
Product Rulelogb(xy) = logb(x) + logb(y)Convert to exponents, multiply, convert back
Quotient Rulelogb(x/y) = logb(x) − logb(y)Convert to exponents, divide, convert back
Power Rulelogb(xn) = n · logb(x)Convert to exponents, apply exponent to exponent
Change of Baseloga(x) = logb(x) / logb(a)Take log of both sides of definition
Reciprocallogb(a) = 1 / loga(b)Solve the definition equation for the target

How to Practice Effectively

Don't just read proofs—reproduce them from scratch. Here's what works:

One properly worked proof teaches you more than ten half-finished attempts. Quality over quantity here.

If you're still struggling, go back to the definition: logb(x) = y means by = x. That's the foundation everything else builds on. Master that relationship and every proof becomes manageable.