1)

Here are my stepsGiven y^{2}- 1 = x^{2 }

Show that log_{(y-1)}x + log_{(y+1)}x = 2(log_{(y-1)}x)(log_{(y+1)}x)

Please let me know if I make any mistakes

log_{(y-1)(y+1)}[(y-1)(y+1)]= log_{(y-1)(y+1)}x^{2}

log_{(y-1)(y+1)}[(y-1)(y+1)]= 2 log_{(y-1)(y+1)}x

log_{(y-1)}[(y-1)(y+1)] / log_{(y+1)}[(y-1)(y+1)] = 2 ( log_{(y-1)(y+1)}x )

log_{(y-1)}(y+1) / log_{(y+1)}(y-1) = 2 ( log_{(y-1)(y+1)}x )

I got stuck here. I am not sure how to move forward from here.

2)Here are my stepsFind the value of x such that (2/x)^{logn4}- (3/x)^{logn9}= 0

x^{2}- 2y^{2}= 2x - 3y

log_{n}4 [ log_{n}(2/x) ] = log_{n}9 [ log_{n}(3/x) ]

log_{n}2^{2}[ log_{n}(2/x) ] = log_{n}3^{2}[ log_{n}(3/x) ]

log_{n}2^{2}[ log_{n}(2) - log_{n}(x)] = log_{n}3^{2}[ log_{n}(3) - log_{n}(x) ]

log_{n}2^{2}log_{n}(2) - log_{n}2^{2}log_{n}(x) = log_{n}3^{2 }log_{n}(3)- log_{n}3^{2 }log_{n}(x)

I got stuck here and I am not sure how to move forward from here.

3)Here are my following stepsIf 2^{x}3^{y}= 3^{x}4^{y}= 6

show by taking logarithms to base 2 and eliminating log_{2}3, that

x^{2}- 2y^{2}= 2x - 3y

2^{x}3^{y}= 3^{x}4^{y}= 6

2^{x}/3^{x}= 4^{y}/3^{y}

(2/3)^{x}= (4/3)^{y}

x log_{2}(2/3) = y log_{2}(4/3)

x [ log_{2}(2) - log_{2}(3) ] = y [ log_{2}(4) - log_{2}(3) ]

x log_{2}(2) - 2y log_{2}(2) = x log_{2}(3) - y log_{2}(3)

x - 2y = log_{2}(3) [(x-y)]

log_{2}(3) = (x - y)/(x - 2y)

I know I did something wrong here but I just couldn't figure out.

Thank you very much. Hope someone can help me out! I will appreciate it. I don't need a full solution but I would like some hints on how should I carry on. Thank you!!