find the value of y such that
4y+1=82y-1

the y+1 is a power and so is 2y-1
I don't know how to write it in that format
I know the answer I just don't understand how you arrive at it
its 5/4

Just use the ^ key. 4^(y+1)=8^(2y-1)

That's what that key is for.

I will use LaTex though.

4^{y+1}=8^{2y-1}

ln of both sides:

(y+1)ln(4)=(2y-1)ln(8)

(y+1)ln(2^{2})=(2y-1)ln(2^{3})

2(y+1)ln(2)=3(2y-1)ln(2)

2(y+1)=3(2y-1)

Solve for y.

Most people use a "^" to indicate a power, sort of like this:
4^(y+1) = 82^(y-1)

I'll do it differently:

4^{(y+1)}\,=\,82^{(y-1)}

If you take the logarithm of both sides, you'll get

log(4^{(y+1)})\,=\,log(82^{(y-1)})

Note this:

log(A^x) = x\,log(A)

So using that we have

(y+1)\,log(4)\,=\,(y-1)\,log(82)

you can now rearrange it and solve it. To simplify that, let A = log(4) and B = log(82):

A(y+1)\, = \, B (y-1)

Ay + A\, = \, By - B

(A-B)y\, =\, -(A+B)

y\, =\, -\frac{(A+B)}{A-B}

log(4) = 0.602
log(82) = 1.9138
A+B = 2.516
A-B = -1.3118

y\,=\,-\frac{(A+B)}{A-B}\,=\,1.918

I don't get 5/4 = 1.25. Check me and see if I did something wrong.

By the way, you don't have to use common logs. You can use ln or logarithms to any base.

I see that Galactus posted an answer while I was working on it and he (correctly) used 8^(2y-1) instead of 82^(y-1). That's why I got the "wrong" answer. He also used natural logs and he recognized that 2 is the cube root of 8.