Question

Find the value of x in the following:

log (3x + 2) + log (3x – 2) = 1 + log 2 + log 7

Answer 1

Given: log (3x + 2) + log (3x – 2) = 1 + log 2 + log 7

Step-wise calculation:

1. Use log a + log b = log (ab): 

Left = log [(3x + 2)(3x – 2)]

2. Combine right:

log 2 + log 7 = log 14

And 1 = log 10 (base 10)

So right = log 10 + log 14 

= log (10 × 14) 

= log 140

3. Equate arguments:

(3x + 2)(3x – 2) = 140

4. Expand/use difference of squares:

9x² – 4 = 140 

⇒ 9x² = 144

⇒ x² = 16

⇒ x = ±4

5. Check domain:

3x – 2 > 0 and 3x + 2 > 0 require `x > 2/3`

So x = –4 is invalid.

Only x = 4 satisfies the log domain.