`log_3(x) + log_3(x - 2) = 1` Solve for x or b

`log_3(x) + log_3(x - 2) = 1`

The logarithms at the left side have the same base. So express the left side with one logarithm only using the rule `log_b (M) + log_b (N) = log_b(M*N` ).

`log_3(x * (x-2)) = 1`

`log_3 (x^2 - 2x) = 1`

Then, convert this to exponential form.

Take note that if a logarithmic equation is in the form

`y = log_b (x)`

its equivalent exponential equation is

`x...

`log_3(x) + log_3(x - 2) = 1`

The logarithms at the left side have the same base. So express the left side with one logarithm only using the rule `log_b (M) + log_b (N) = log_b(M*N` ).

`log_3(x * (x-2)) = 1`

`log_3 (x^2 - 2x) = 1`

Then, convert this to exponential form.

Take note that if a logarithmic equation is in the form

`y = log_b (x)`

its equivalent exponential equation is

`x = b^y`

So converting

`log_3 (x^2 - 2x) =1`

to exponential equation, it becomes:

`x^2-2x = 3^1`

`x^2 - 2x = 3`

Now the equation is in quadratic form. To solve it, one side should be zero.

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

Factor the left side.

`(x - 3)(x +1)=0`

Set each factor equal to zero. And isolate the x.

`x - 3 = 0`

`x=3`

`x + 1=0`

`x = -1`

Now that the values of x are known, consider the condition in a logarithm. The argument of a logarithm should always be positive.

In the equation

`log_3(x) + log_3(x - 2)=1`

the arguments are x and x - 2. So the values of these two should all be above zero.

`x gt 0`

`x - 2gt0`

Between the two values of x that we got, it is only x = 3 that satisfy this condition.

Therefore, the solution is `x=2` .