Question 718243
When your variable is in an exponent, the general rule is that you use logarithms to solve for the variable. For example, if we have a power (remember your definition of a power?) equal to a number as 
(1) 2^x = 3
The procedure is to take the LN of both sides of the equation and get
(2) LN(2^x) = LN(3)
Now we apply the rule that the LN of a power is equal to the exponent times the LN of the base and get
(3) x*LN(2) = LN(3)
Now you treat the LN(2) and LN(3) just as you would we any real number and get
(4) x = LN(3)/LN(2)
Now use your scienfific calculator to get
(5) x = 1.0986.../0.6931... or
(6) x = 1.584...
Rarely will it be a "nice" round number, so keep all intermediate computational values in the calculator; rounding only the final answer.
So much for the lesson, now let's do your problem
(7) {{{5^(6x+3) = 37}}}
Note we have a power on the left and a real number on the right just as we do in example (1).
Now take the LN of (7) and get
(8) {{{(6x+3)*LN(5) = LN(37)}}} or
(9) {{{(6x+3) = LN(37)/LN(5)}}} or
(10) {{{6x = (LN(37)/LN(5))-3}}} or
(11) {{{x = ((LN(37)/LN(5))-3)/6}}}  or
(12) x = (((3.6109...)/(1.6094...))-3)/6 or
(13) x = ((2.243...)-3)/6 or
(14) x = (-0.7564...)/6 or
(15) x = -0.126...
Let's check this using (7).
Is (5^(6*(-0.126...)+3) = 37)?
Is (5^(2.243...) = 37)?
Is (37 = 37)? Yes
Answer: x is approximately -0.1260684
PS In the above calculations you could use LOG instead of LN, it doen't matter.