# What is the answer for question 6. b) ? http://postimg.org/image/xusvq33e3/

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You may move all the logarithms to one side, such that:

`log_7 x + log_7 (x - 1) - log_7 2x = 0`

Since the logarithms both sides share the same base` 7` , you may use the following logarithmic identities, such that:

`log x + log y = log(xy)`

`log x - log y = log(x/y)`

Reasoning by analogy yields:

`log_7 x + log_7 (x - 1) = log_7 x(x - 1)`

`log_7 x(x - 1) - log_7 2x = log_7 x(x - 1)/(2x)`

Replacing `log_7 x(x - 1)/(2x)` for log_7 x + log_7 (x - 1) - log_7 2x yields:

`log_7 x(x - 1)/(2x) = 0 => x(x - 1)/(2x) = 7^0`

`x^2 - x = 2x => x^2 - 3x = 0 `

Factoring out `x` yields:

`x(x - 3) = 0`

Using zero product property yields:

`x = 0`

`x - 3 = 0= > x = 3`

Testing the values `x = 0` and `x = 3 ` in equation yields:

`log_7 0 + log_7 (0 - 1) - log_7 2*0` invalid

`log_7 3 + log_7 (3 - 1) - log_7 2*3 = 0`

`log_7 6 - log_7 6 = 0` valid

**Hence, evaluating the solution to the given logarithmic equation, yields `x = 3.` **

`log_(7)x+log_(7)(x-1)=log_(7)2x` (given)

Using the logarithm rule of `log_(b)a+log_(b)c=log_(b)ac`

`log_(7)x(x-1)=log_(7)2x`

`rArr x(x-1)=2x`

`rArr x^2-x=2x`

`rArr x^2-3x=0`

`rArr x(x-3)=0`

The value of x cannot be 0 because log 0 is undefined.

**Therefore, x=3 is the required answer.**