log5(7)+log5(3)

$$\\log_5(7)+log_5(3)\\\\
=log_521\\\\
=\frac{log_{10}21}{log_{10}5}\\\\
=1.8917\qquad $correct to 4 decimal places$$$

.

Logs are exponents.

You add exponents when you have a multiplication problem with the same bases.

So, if you have added logs, it was a multiplication problem.

log7 + log3  =  log21   (for every base)

Do you need a decimal answer?  If you do, use the change-of-base formula. Any question? Post back.