C1 paper question help?

Can anyone help me with part b) of this question? thanks!

Can you do it after noting

$4^{x+1}=4 \times 4^x$

$2^{x} \times 4 \times 4^x = 4 \times 8^{x}$

No can you please run through the rest of the method? thanks for your time.

$2^{x} \times 4 \times 4^x = 4 \times 8^{x}$

$4 \times 8^x = 8$

$8^x = 2$

$x=\log_8 2$

I'm sure you can finish it from there.

I dont understand how you arrived at your second line of working looking at your first? Need to revise this topic

If you still need help, Without using logarithms because that's C2 you do;
Not sure how to do maths thing but try to read it,
(2^x) X (4^x+1) = 8
(2^x) X ((2^2)^x+1) = 2^3
so simplify then add the powers: (2^x) X (2^2x+2) = 2^3 => (2^3x+2) = 2^3 => 3x+2=3 => 3x=1 => x=1/3

I don't think that's valid working.

I don't think that's valid working.

There's no rule that states that $a^n \times b^n=(a \times b)^n$

Where is it invalid?

Edit: Unless you missed the context where I was considering the LHS only for the first line?

....

It's valid, it's just that it's using logarithms and that's in C2

2^2 x 4^2 = 64
8^2 = 64

If you still need help, Without using logarithms because that's C2 you do;
Not sure how to do maths thing but try to read it,
(2^x) X (4^x+1) = 8
(2^x) X ((2^2)^x+1) = 2^3
so simplify then add the powers: (2^x) X (2^2x+2) = 2^3 => (2^3x+2) = 2^3 => 3x+2=3 => 3x=1 => x=1/3