**Solving Logarithmic Equations with Radicals :**

Here we are going to see how to solve logarithmic equations with radicals.

**Product rule :**

log _{a} (m ⋅ n) = log_{a}m + log_{a}n

**Quotient rule :**

log _{a} (m / n) = log_{a}m - log_{a}n

**Power rule :**

log _{a} m^{n} = n log_{a}m

**Change of base rule :**

log_{b}a = log_{x}a ⋅ log_{b}x

log_{b}a = log_{x}a / log_{x}b

**Example 1 :**

Solve the following equation :

log_{5} √(7x - 4) - 1/2 = log_{5} √(x + 2)

**Solution :**

log_{5} √(7x - 4) - 1/2 = log_{5} √(x + 2)

Subtract log_{5} √(x + 2) from each side.

log_{5} √(7x - 4) - log_{5} √(x + 2) - 1/2 = 0

Add 1/2 to each side.

log_{5} √(7x - 4) - log_{5} √(x + 2) = 1/2

Use quotient rule.

log_{5}[√(7x - 4) / √(x + 2)] = 1/2

Convert to exponential form.

[√(7x - 4) / √(x + 2)] = 5^{1/2 }

(7x - 4) / (x + 2) = √5^{ }

Square each side.

(7x - 4) / (x + 2) = 5^{ }

Multiply each side by (x + 2).

7x - 4 = 5 (x + 2)

7x - 4 = 5x + 10

Subtract 5x from each side.

2x - 4 = 10

Add 4 to each side.

2x = 14

Divide each side by 2.

x = 7

So, the value of x is 7.

**Example 2 :**

Solve the following equation

log_{3} √(5x - 2) - 1/2 = log_{3} √(x + 4)

**Solution :**

log_{3} √(5x - 2) - (1/2) = log_{3} √(x + 4)

Subtract log_{5} √(x + 4) from each side.

log_{3} √(5x - 2) - log_{3} √(x + 4) - 1/2 = 0

Add 1/2 to each side.

log_{3} √(5x - 2) - log_{3} √(x + 4) = 1/2

Use quotient rule.

log_{3}[√(5x - 2) / √(x + 4)] = 1/2

Convert to exponential form.

[√(5x - 2) / √(x + 4)] = 3^{1/2}

(5x - 2) / (x + 4) = √3^{ }

Square each side.

(5x - 2) / (x + 4) = 3^{ }

Multiply each side by (x + 4).

5x - 2 = 3 (x + 4)

5x - 2 = 3x + 12

Subtract 3x from each side.

2x - 2 = 12

Add 2 to each side.

2x = 14

Divide each side by 2.

x = 7

So, the value of x is 7.

After having gone through the stuff given above, we hope that the students would have understood how to solve logarithmic equations with radicals.

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