Question

What is the value of x in the expression log(x^2-9)=log7+log(x+3)?

Answer 1

solution :

log(x²-9) = log7+log(x+3)

⇒ log(x+3)(x-3) = log7+log(x+3)

⇒ log(x+3)+log(x-3) = log7+log(x+3)

⇒ log(x-3)-log7 = log(x+3)-log(x+3)

⇒ $log(\frac{x-3}{7} ) = 0$

⇒ $\frac{x-3}{7} = e^{0}$

⇒ x-3 = 7

⇒ x= 10

Answer 2

Value of x is 10  if log(x² - 9) = log 7 + log (x - 3)

Given:

log(x² - 9) = log 7 + log (x - 3)

To Find:

Value of x

Solution:

log (ab) = log a + log b

a² - b² = (a + b)(a - b)

Antilog ( log a) = a

Step 1:

Using identity a² - b² = (a + b)(a - b) where a = x  b = 3

log(x² - 9) = log 7 + log (x - 3)

=> log(x² -3²) = log 7 + log (x - 3)

=> log ((x + 3)(x - 3)) = log 7 + log (x + 3)

Step 2:

Using identity log (ab) = log a + log b where a = x + 3 , b= x - 3

 log  (x + 3) + log(x - 3) = log 7 + log (x + 3)

=>   log  (x - 3)  =  log 7    cancelling  log (x+ 3) from both sides

Step 3:

Taking antilog both sides

x - 3 = 7

=> x = 10

Value of x is 10  if log(x² - 9) = log 7 + log (x - 3)