Solve for x: log 3 (7x+5)=2?

Assumptions: Not knowing how this equation is formulated if parentheses were added to make it clearer:
1.The equation is: ln(2*(7*x+5))=2
x=(e^2-15)/32 = -0.3624259
x = -0.3624259


2. ln(3)*(7*x+4)=3
x=-(4 ln(3) - 3)/(7*ln(3))
x = -0.961531097

Hope I got the right eqn.

log 3 (7x+5)=2
3^2 = 7x + 5
9 = 7x + 5
4 = 7x
x = 4/7

Is this log to base 3 ?

If so , let log be log to base 3

log ( 7 x + 5 ) = 2

7 x + 5 = 3 ²

7 x = 4

x = 4 / 7

log_3(7x + 5) = 2
7x + 5 = 3^2
7x = 9 - 5
x = 4/7

3(7x+5)=2
21x+15=2
-15 -15
21x =-17
/21 /21
x =-0.8

please be clear with the question. is the 3 as the base or is it in multiplication with 7x+5. if it is the base then here is the answer

take anti log on both the sides ao u get

7x+5 = 3^2
7x +5 = 9
x = 4/9