Logarithmic equations practice worksheet with worked example and student exercises.

Let’s solve each equation one by one, step by step.

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Equation 1:
> log₂(5) + log₂(y) = 4

We use the logarithm rule:
logₐ(A) + logₐ(B) = logₐ(A × B)

So:
→ log₂(5 × y) = 4
→ log₂(5y) = 4

Now convert from log form to exponential form:
If log₂(5y) = 4, then
→ 2⁴ = 5y
→ 16 = 5y
→ y = 16/5
→ y = 3.2

✔ Check: Plug back in — log₂(5) + log₂(3.2) = log₂(5×3.2) = log₂(16) = 4 → Correct!

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Equation 2:
> log₃(x + 8) - log₃(x - 3) = 3

Use the rule:
logₐ(A) - logₐ(B) = logₐ(A/B)

So:
→ log₃[(x + 8)/(x - 3)] = 3

Convert to exponential form:
→ (x + 8)/(x - 3) = 3³ = 27

Now solve for x:

Multiply both sides by (x - 3):
→ x + 8 = 27(x - 3)

Expand right side:
→ x + 8 = 27x - 81

Bring all terms to one side:
→ 8 + 81 = 27x - x
→ 89 = 26x
→ x = 89/26

Check if this is valid:
Denominator was (x - 3). If x = 89/26 ≈ 3.42, then x - 3 ≈ 0.42 > 0 → OK.
Also, x + 8 > 0 → obviously true.

✔ So x = 89/26 is acceptable.

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Equation 3:
> 2log₃(x + 2) = log₃(2x + 1) + 1

First, use power rule on left:
a·logₐ(B) = logₐ(Bᵃ) → so 2log₃(x+2) = log₃((x+2)²)

Right side: we have “+1”. Remember that 1 = log₃(3), because log₃(3) = 1.

So rewrite equation:

→ log₃((x + 2)²) = log₃(2x + 1) + log₃(3)

Now combine right side using addition rule:
→ log₃((x + 2)²) = log₃[ (2x + 1) × 3 ]

Since logs are equal and same base, arguments must be equal:

→ (x + 2)² = 3(2x + 1)

Expand both sides:

Left: x² + 4x + 4
Right: 6x + 3

Set equal:

→ x² + 4x + 4 = 6x + 3

Bring all to left:

→ x² + 4x + 4 - 6x - 3 = 0
→ x² - 2x + 1 = 0

Factor:

→ (x - 1)² = 0
→ x = 1

Check domain:
Original logs:
- log₃(x + 2) → x + 2 = 3 > 0 → OK
- log₃(2x + 1) → 2(1)+1=3 > 0 → OK

✔ Valid solution: x = 1

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Final Answer:
y = 16/5, x = 89/26, x = 1