Solve : Log5( X + 1 ) - 1 = 1 + Log5( X - 1 ).

Solve : log₅( x + 1 ) - 1 = 1 + log₅( x - 1 ).

Sum

log₅( x + 1 ) - 1 = 1 + log₅( x - 1 )

⇒ log₅( x + 1 ) - log₅( x - 1 ) = 2

⇒ `log_5 ( x + 1 )/( x - 1 ) = 2`

⇒ `( x + 1 )/( x - 1 ) = 5^2`

⇒ `( x + 1 )/( x - 1 ) = 25`
⇒ x + 1 = 25( x - 1 )
⇒ x + 1 = 25x - 25
⇒ 25x - x = 25 + 1
⇒ 24x = 26
⇒ x = `26/24 = 13/12`.

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More About Logarithm

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Chapter 8: Logarithms - Exercise 8 (D) [Page 111]

Chapter 8 Logarithms
Exercise 8 (D) | Q 14 | Page 111

If x = 1 + log 2 - log 5, y = 2 log3 and z = log a - log 5; find the value of a if x + y = 2z.

If x = log 0.6; y = log 1.25 and z = log 3 - 2 log 2, find the values of :
(i) x+y- z
(ii) 5^{x + y - z}

If a² + b² = 23ab, show that:

log `(a + b)/5 = 1/2`(log a + log b).

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Find x and y, if `("log"x)/("log"5) = ("log"36)/("log"6) = ("log"64)/("log"y)`

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If a = `"log" 3/5, "b" = "log" 5/4 and "c" = 2 "log" sqrt(3/4`, prove that 5^a+b-c = 1

Express the following in a form free from logarithm:
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