I need to isolate n in this equation

i) Cross multiplying, G*r = y - y*{(1 + r)^(-n)}

ii) G*r - y = - y*{(1 + r)^(-n)}

iii) ==> {(1 + r)^(-n)} = -(G*r - y)/y

iv) Taking natural log on both sides,

(-n)*ln(1 + r) = ln{-(G*r - y)/y}

So, -n = ln{-(G*r - y)/y}/ln(1 + r)

n = -ln{-(G*r - y)/y}/ln(1 + r) - Which is your requirement.

However, the same can be expressed in much better form as:

n = [ln|y| - ln|y - Gr|]/[ln|1 + r|]

G=y*(1-(1+r)^-n)/r or
G*r = y*(1-(1+r)^-n) or
{(G*r)/y} = 1 - (1+r)^-n or
(1+r)^-n = 1 - {(G*r)/y} or
(1+r)^n = 1/[1 - {(G*r)/y}] or
n*ln(1+r) = ln1 -ln[1 - {(G*r)/y}] or
n = -ln[1 - {(G*r)/y}]/{ln(1+r)}
There is some slip of error in the expression given as answer.

yust proceed step by step !

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

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

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

-n*ln(1+r) = ln[ -(Gr-y)/y ]

n = -ln[ -(Gr-y)/y ] / ln(1+r)
---------------------------------

G=y*(1-(1+r)^-n)/r
r*G/y = 1-(1+r)^-n
(1+r)^-n = 1 - r*G/y
(1+r)^-n = (y- r*G)/y
taking Log
-nLog(1+r) = Log [(y- r*G)/y]
n = - Log [(y- r*G)/y] / Log(1+r)
n = -Log[-(r*G - y)/y] / Log(1+r)