**Solving logarithmic equations for x :**

Here we are going to see how to solve logarithmic equations for a particular variable.

**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_{4} (x + 4) + log_{4} 8 = 2

**Solution :**

log_{4} (x + 4) + log_{4} 8 = 2

Let us combine the two terms on the left side

log_{4} [8 ⋅ (x + 4)] = 2

log_{4} (8 x + 32) = 2

8x + 32 = 4^{2}

8x + 32 = 16

Subtract by 32 on both sides

8x = 16 - 32

8x = -16

Divide by 8 on both sides

x = -16/8

x = -2

Hence the value of x is 8.

**Example 2 :**

Solve the following equation :

log_{6} (x + 4) - log_{6} (x - 1) = 1

**Solution :**

log_{6} (x + 4) - log_{6} (x - 1) = 1

Let us combine the two terms on the left side

log_{6} [(x + 4) / (x - 1)] = 1

[(x + 4) / (x - 1)] = 6^{1}

(x + 4) / (x - 1) = 6

x + 4 = 6(x - 1)

x + 4 = 6x - 6

Subtract both sides by 6x

x - 6x + 4 = -6

-5x + 4 = -6

Subtract both sides by 4

-5x = -6 - 4

-5x = -10

Divide both sides by -5

x = -10/(-5)

= 2

Hence the value of x is 2.

**Example 3 :**

Solve the following equation :

log_{2} x + log_{4} x + log_{8} x = 11/6

**Solution :**

In order to change the bases same, we are going to convert

log_{2} x + log_{4} x + log_{8} x = 11/6

(1/log_{x} 2) + (1/log_{x} 4) + (1/log_{x} 8) = 11/6

(1/log_{x} 2) + (1/log_{x} 2^{2}) + (1/log_{x} 2^{3}) = 11/6

(1/log_{x} 2) + (1/2log_{x} 2) + (1/3log_{x} 2) = 11/6

(1/log_{x} 2) (1 + 1/2 + 1/3) = 11/6

(1/log_{x} 2) (11/6) = 11/6

1/log_{x} 2 = 1

1 = log_{x} 2

x = 2

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

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