Find the values of x that satisfy the logarithmic equation involving base-5 and base-4 logarithms | Step-by-Step Solution

What We're Solving:

We need to find the value(s) of x that make log₅(x) × log₄(27-x) = 1 true, where x must be between 0 and 27.

The Approach:

This is a tricky logarithmic equation because we have two different bases (5 and 4) multiplied together. Our strategy will be to use the change of base formula to convert everything to the same base (natural log works great), then solve the resulting equation. We'll also need to be careful about the domain - both arguments of our logarithms must be positive.

Step-by-Step Solution:

Step 1: Check our domain restrictions For log₅(x) to exist: x > 0 For log₄(27-x) to exist: 27-x > 0, so x < 27 This confirms our domain: 0 < x < 27 ✓

Step 2: Apply change of base formula Remember: log_a(b) = ln(b)/ln(a)

So our equation becomes: [ln(x)/ln(5)] × [ln(27-x)/ln(4)] = 1

Step 3: Simplify the equation ln(x) × ln(27-x) = ln(5) × ln(4)

Let's call ln(x) = u and ln(27-x) = v for easier handling: u × v = ln(5) × ln(4)

Step 4: Make a substitution Since x = 5ᵗ for some value t. Then: log₅(x) = log₅(5ᵗ) = t

Our equation becomes: t × log₄(27-5ᵗ) = 1 So: log₄(27-5ᵗ) = 1/t

Step 5: Convert back from logarithmic form If log₄(27-5ᵗ) = 1/t, then: 27-5ᵗ = 4^(1/t)

Step 6: Try strategic values If t = 2, then x = 5² = 25 We need: 27-25 = 2, and 4^(1/2) = 2 ✓ This works! Let's verify: log₅(25) × log₄(2) = 2 × (1/2) = 1 ✓

The Answer:

x = 25

Let's double-check: log₅(25) = 2 (since 5² = 25) and log₄(2) = 1/2 (since 4^(1/2) = 2) Therefore: 2 × (1/2) = 1 ✓

Memory Tip:

When dealing with logarithmic equations with different bases, the change of base formula is your best friend! Also, when equations look complex, try substituting the variable with exponential forms - sometimes strategic guessing with simple values (like 1, 2, 1/2) can lead you to the solution faster than pure algebra. Always verify your answer by plugging it back into the original equation!

Great work tackling this challenging problem! Logarithmic equations can be tricky, but breaking them down step by step makes them much more manageable. 🌟