1) Solver for x:

log7(x+4) + log7(x-2)=1

First, recall the rule: Log(a*b) = log(a) + log(b). So the first step in your problem is to combine the two log terms - here I assume that when you wrote log7(x+4) you meant:

log _7 (x+4), correct?

So:

log(x+4) + log(x-2) = 1\\
log((x+4)*(x-2)) = log(x^2 +2x-8) = 1


Now raise both sides by the power of 7 (the base of the log), recalling that

7 ^ {log_7 a} = a
\\

7 ^ {log _7 (x^2 +2x -8)} = 1\\
x^2 +2x -8 = 7^1 = 7


Can you take it from here?

1) Solver for x:

log7(x+4) + log7(x-2)=1

7x+4+7x-14=1
14x=1+14-4
14x=11
x=1.3