Question

2 log(a+b)+log(a-b)-log(a2-b2)=log x,find x

Answer 1

Step-by-step explanation:

Stepwise solution is provided in enclosure.

Answer 2

Answer: x=(a+b)/(a-b)

Step-by-step explanation: Given: $2 log(a+b)+log(a-b)-log(a2-b2)=log x$

Now first have a look at the properties of log functions

$nlogm=logm^{n}\\ logm+logn=log(mn)\\logm-logn=log(m/n)$

Therefore;

$2 log(a+b)+log(a-b)-log(a2-b2)=log x$  can be written as

∴$log(a+b)^{2}+log(a-b)-log(a^{2}-b^{2})$

∴$log{(a+b)^{2}(a-b)}-log(a^{2}-b^{2} )=logx$

∴$log(a+b)^{2}(a-b)/(a^{2}-b^{2})=logx$

Removing log from both the sides we get,

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

∴$x=(a+b)(a+b)(a-b)/(a+b)(a-b)$

therefore, on dividing the same parts we get,

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

Hence the required answer is x=(a+b)/(a-b)

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