what is the value of the expression log(x^2-9)=log 7+ log(x+3)
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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)
⇒
⇒
⇒ x-3 = 7
⇒ x= 10
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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)
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