Question

Log(x-1)+log(x+1) = 3log^2+ log3

Answer 1

Answer:

$x = + \: 5$

Step-by-step explanation:

Using Logarithmic properties,

For LHS

$log(x - 1) + log(x + 1) = log((x - 1)(x + 1)) = log(x^{2} - 1 )$

Similarly for RHS,

$3 log(2) = log(2^{3} ) = log(8)$

And

$log(8) + log(3) = log(8 \times 3) = log(24)$

Now equating LHS and RHS, we get

$log(x ^{2} - 1 ) = log(24) \\ x ^{2} - 1 = 24 \\ x^{2} = 25 \\ x = + \: or \: - 5$

But when we plug in - 5 in the original equation, we turn the question to log(-6)+log(-4)=RHS.

Since logarithms cannot have negative numbers inside them, hence

x=5