log6 (x+ 41) - log6 (x+1) = 2

First we will find the domain.

==> x+ 41>0 ==> x > -41

==> x + 1 > 0 ==> x > -1

==> Then the domain is x > -1 .....(1)

We will use the logarithm properties to solve for x.

We know that log a - log b= log (a/b)

==> log6 (x+41) / (x+1) =2

Now we will rewrite using the exponent form.

==> (x+41)/(x+1) = 6^2

==> (x+41)/ (x+1) = 36

Multiply by x+1

==> x+ 41 = 36(x+1)

==> x+ 41 = 36x + 36

==> 35x = 5

=> x = 5/35 = 1/7

==> x= 1/7 > -1 ( then the solution within the domain.)

**Then the solution is x= 1/7**

We have to solve: log(6) ( x+41 ) - log(6) (x+1) = 2

use the property log a + log b = log(a*b)

log(6) ( x+41 ) - log(6) (x+1) = 2

=> log(6)[( x + 41)/(x + 1)] = 2

=> (x + 41) / (x + 1) = 6^2

=> (x + 41) / (x + 1) = 36

=> x + 41 = 36x + 36

=> 35x = 5

=> x = 1/7

we see that log(6) ( x+41 ) and log(6) (x+1) are defined for x = 1/7.

**Therefore the solution for the equation is x = 1/7.**

We notice that the bases of logarithms are matching, so we could transform the difference into a quotient.

log(6)(x+41) - log(6)(x+1) = log(6) [(x+41)/(x+1)]

But log(6)(x+41) - log(6)(x+1) = 2 => log(6) [(x+41)/(x+1)] = 2

We'll take antilogarithm and we'll get:

[(x+41)/(x+1)] = 6^2

[(x+41)/(x+1)] = 36

x + 41 = 36x + 36

We'll isolate x to the left:

x - 36x = 36 - 41

-35x = -5

x = 5/35

x = 1/7

The constraints of existence of logarithms gives the interval of possible values for x: (-1 ; +infinite).

**Since x = 1/7 belongs to the range (-1 ; +infinite), then x = 1/7 is the solution of the equation.**