solve 4^2x+1 = 8 .....mathematics help?

There are two possibilities
1)
4^(2x) + 1 = 8
Add -1 both sides
4^(2x) = 7
Takin log both sides
2x(log4) = log7
x = log7/2*(log4) = 0.8451/1.2041
OR x = 0.7018

2)

4^(2x + 1) = 8
(2^2)^(2x + 1) = 2^3
2^[2(2x + 1)] = 2^3
2(2x + 1) = 3
2*2x + 2*1 = 3
4x = 3 - 2
x = 1/4
OR x = 0.25

4^(2x+1) = 8 (put log at both side so that you can solve it)

log 4^(2X+1) = log 8

(2x+1) log 4 = log 8

2x + 1 = log 8 / log 4 = 1.5

2x + 1 = 1.5

2x = 1.5 -1 = 0.5

x = 0.5 / 2 = 0.25

Question is unclear.

Will assume that you mean :-

4^(2x + 1 ) = 8

2^(4x + 2) = 2^3

4x + 2 = 3

4x = 1

x = 1/4

The other answers are right if its 4^(2x+1).
But if its 4^(2x) +1, they are not. So be careful about typing the correct problem when you post a difficulty here.

4^(2x + 1) = 8
(2^2)^(2x + 1) = 2^3
2^[2(2x + 1)] = 2^3
2(2x + 1) = 3
2*2x + 2*1 = 3
4x = 3 - 2
x = 1/4 (0.25)

4^2x+1 = 8

We know that 4 = 2^2 and 8 = 2^3 so

2^2(2x+1) = 2^3

Therefore, by equating the powers

2(2x+1) = 3

2x+1 =3/2
2x = 3/2 -1 = 1/2
x=1/4

If the question is

4^(2x+1) = 8 then this is how you go

We know 4 = 2^2 and 8 = 2^3 so

2^2(2x+1) = 2^3

Therefore

2(2x+1) = 3

2x+1 =3/2
2x = 3/2 -1 = 1/2
x=1/4

4^2x+1 = 8 => 2^(4x+2) = 2^3 => 4x+2 = 3 => x = 1/4.