Solution

Use the definition of logarithm: \[ \solve{ 5^{3x}&=&35\\ 3x&=&\log_5(35)\\ x&=&\frac{\log_5(35)}{3} } \] Now, this solution will be valid in most cases, but you should always keep an eye for possible simplifications. Logarithms have a lot of properties that allow for alternate ways to rewrite the answers. In this case: \[ \solve{ \frac{\log_5(35)}{3}&=&\frac{\log_5(5\times 7)}{3}\\ &=&\frac{\log_5(5)+\log_5(7)}{3}\\ &=&\frac{1+\log_5(7)}{3}\\ &=&\frac{1}{3}+\frac{1}{3}\log_5(7) } \] I also prefer to write logarithms with the fraction in front, when possible, since the power rule allows for yet more ways to play around with rewriting the answer. For now, I will leave our final answer as above.