Question

if log(a-b)-log(a+b)=logb/a, find a^2/b^2+b^2/a^2

Answer 1

loga²/b²+logb²/a²
loga/b=loga-logb
loga²-logb²+logb²-loga²=0

Answer 2

Answer:

6

Step-by-step explanation:

$log(a-b) - log(a+b)=log(b/a)$

$log((a-b)/(a+b))=log(b/a)$

Taking Antilog both side

$(a-b)/(a+b) =(b/a)$

$a(a-b)=b(a+b)$

$a^{2} -ab=b^{2}+ab$

$a^{2}-b^{2}=2ab$

Squaring both side

$a^{4}+b^{4}-2a^{2}b^{2} = 4a^{2}b^{2}$

$a^{4} + b^{4} = 6a^{2}b^{2}$

$(a^{4}+b^{4})/a^{2}b^{2} = 6$     ----(i)

Given

$a^{2}/b^{2} + b^{2}/a^{2} = (a^{4}+b^{4})/a^{2}b^{2}$    ----(ii)

From (i) and (ii)

$a^{2}/b^{2} + b^{2}/a^{2} = 6$