# log(5x+1) - log(x+1) = log(2x-1)

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### 1 Answer

You need to solve for x the given logarithmic equation and you need to perform the first step converting the difference of logarithms from the left, into the logarithm of quotient, using the following logarithmic identity, such that:

`log a - log b = log (a/b)`

Reasoning by analogy, yields:

`log (5x + 1) - log (x + 1) = log ((5x + 1)/(x + 1))`

Replacing `log ((5x + 1)/(x + 1))` for `log (5x + 1) - log (x + 1) ` yields:

`log ((5x + 1)/(x + 1)) = log (2x - 1)`

Equating the arguments of logarithms yields:

`((5x + 1)/(x + 1)) = (2x - 1) `

`((5x + 1)/(x + 1)) - (2x - 1) = 0`

Bringing the terms to a common denominator yields:

`(5x + 1 - (2x - 1)(x + 1))/(x + 1) = 0`

Since` (x + 1) != 0 => 5x + 1 - (2x - 1)(x + 1) = 0`

`5x + 1 - 2x^2 - x + 1 = 0 => 2x^2 - 4x - 2 = 0 => x^2 - 2x - 1 = 0`

You may use quadratic formula such that:

`x_(1,2) = (2+-sqrt(4 + 4))/2 => x_(1,2) = (2+-2sqrt2)/2`

`x_(1,2) = (1+-sqrt2)`

You should notice that each logarithm holds for an interval of values of x, such that:

`log(5x+1) EE if 5x + 1 > 0 => x > -1/5`

`log(x+1) EE if x + 1 > 0 => x > -1`

`log(2x-1) EE if 2x - 1 > 0 => x > 1/2 > -1/5 > -1`

You need to notice that only `x = (1+sqrt2) > 1/2.`

**Hence, evaluating the solution to the given equation yields **`x = (1+sqrt2).`