Solve
logbase4(x+3)+logbase4(x-3)=2
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1. x+3 > 0 and x - 3 > 0 ; then, x > - 3 and x > + 3 ; finally, x >  3.
2. logbase4[(x+3)(x-3)] = 2 ; we used the Log Rules:1) logb(mn) = logb(m) + logb(n);
3. We have (x+3)(x-3) = 4^2 because  y = bx  means the exact same thing aslogb(y) = x;
4. x^2 - 9 = 16 because we have the formula (a+b)(a-b)=a^2-b^2 ;
5. x^2 = 25 ;
6. x = - 5 or x = +5;
7. Only, +5 verify the condition from 1., where x > + 3;
8. The solution is +5.

The answer is 5. You can simplify logbase4(x+3) + logbase4(x-3) =2 by simply remembering that logb+logc= log(bc). Logbase4(x+3)(x-3)=2. In these problems, you can simplify logbase4 into an exponent using logc=d into C^D So 4^2 = (x+3)(x-3) 16 = x^2-9 Set it equal to 0, move 16 over to the otherside by subtracting 16 0=x^2-9-16 (x^2-25) Then you factor it and set both problems equal to 0 x+5 =0, x=-5 X-5= 0, x=5 Then you substitute x in the beginning equation to see if the log becomes negative because you can't have a negative log. Since -5-3 is -8, you can cancel out the negative 5, therefore, the answer is 5.

It’s hard to understand what you put.. can you show a picture ?