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katykitz
pls help.. the value of x when 2^(3x + 3) = 2^(3x + 1) + 48 is true a. 1 b. 3/2
2^(3x+3)-64=2^(3x+1)-16 so now assume the equation is true for both LHS = RHS =0 2^(3x+3)=64 x=1; 2^(3x+1)=16 x=1; so the answer will be x=1;
how did u get 64 & 16??pls tell me...
just by randomly looking at the problem.. i'm sorry i cant come up with anything definite 2^6=64 and 2^4=16 and 64-16=48 so.. You could just substitute and arrive at the answer easily...
assume the equation is true for both LHS = RHS =0 LHS: 2^(3x+3)=64 2^(3x+3)= 2^6 implies 3x+3=6 (since base is equal) equating 3x+3=6, we get x=1 similarly try for RHS and give me the answer
k...got it ..thank u..:)