I need to isolate n in this equation

G=y*(1-(1+r)^-n)/r

it should give n=-ln(-(G*r-y)/y)/ln(1+r)

but how do you get there?

you need to know the natural logarithm identity that ln(a^x)=xln(a)
the ln() function allows you to put the exponent out front as a coefficient instead.

G=y[1-(1+r)^-n]/r... first you multiply both sides by r and get
Gr=y[1-(1+r)^-n]... then you divide both sides by y and get
Gr/y=1-(1+r)^-n... now we subtract 1 from both sides. (we subtract y/y on the left. y/y=1
(Gr-y)/y=-(1+r)^-n... now we multiply both sides by -1 to move the negative to the left side
-(Gr-y)/y=(1+r)^-n... now you take the natural logarithm of both sides of the function
ln(-(Gr-y)/y)=ln((1+r)^-n)... now you can move the -n down as a coefficient
ln(-(Gr-y)/y)=-n*ln(1+r)... divide both sides by ln(1+r)
ln(-(Gr-y)/y)/ln(1+r)=-n... now you just multiply both sides by -1 to make n positive
-ln(-(Gr-y)/y)/ln(1+r)=n... and you're done!

G = y×(1-(1+r)^-n)/r

Gr = y×(1-(1+r)^-n) (multiply by r)

Gr/y = 1-(1+r)^-n (divide by y)

(1+r)^-n = 1 - Gr/y (multiply by -1, then add 1 to both sides)

1/(1+r)^n = (y-Gr)/y (take the reciprocal of the left side, and get a common denominator on the right)

(1+r)^n = y/(y - Gr) (invert both fractions)

n = log(base 1+r) [y/(y-Gr)] (use the definition of a logarithm)

n = ln [y/(y-Gr)]/ln (1+r) (change the base to e)

You have isolated n here with a usable log base. But I'll add the steps to get to the given answer.