To solve for x in log(x+7)-log3x= log5.

To solve this logarithmatic equation , we use the property of logarithms:

log a+log b = log ab and loga -logb = loga/b.

Therefore log(x+7)-log3x = log(x+7)/3x

Therefore log(x+7)/3x = log5.

Taking antilog, we get:

(x+7)/3x = 5

(x+7) = 5*3x

x+7 = 15x

7 = 15x-x = 14x

7 = 14x

x = 7/14 =1/2.

x = 1/2.

We'll write the equation

log (x+7) - log 3x = log 5

We could use the quotient property of the logarithms:

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

We'll put a = x+7 and b = 3x and we'll get:

log [(x+7)/3x] = log 5

Since the bases are matching, we'll use the one to one property:

(x+7)/3x = 5

We'll cross multiply:

x+7 = 15x

We'll isolate x to the left side. For this reason, we'll subtract 15 both sides:

-14x = -7

We'll divide by -14 both sides:

x = 7/14

**x = 1/2**

**Since the value of x is positive, the solution of the equation is admissible and it is x = 1/2.**