I do not trust your parentheses
eg:
8^(x) + 3 or 8^(x+3)
or
4^(2x) -1 or 4^(2x-1)
Solve 16(4^2x-1)= 8^x+3
3 answers
if it is
16 [ 4^(2x-1) ] = 8^(x+3)
that is
2^4 [ 2^(4x-2) ] = 2^[3(x+3)]
2^[16x-8] = 2^[3x+9]
or
16-x = 3x+9
16 [ 4^(2x-1) ] = 8^(x+3)
that is
2^4 [ 2^(4x-2) ] = 2^[3(x+3)]
2^[16x-8] = 2^[3x+9]
or
16-x = 3x+9
16x-8 = 3x+9
1 answer