Question

log[x+√x^2-1]+log[x-√x^2-1]=0​

Answer 1

Required Answer:-

Correct Question:

• Prove that log(x + √(x² - 1)) + log(x - √(x² - 1)) = 0

Proof:

Taking LHS,

We know that,

$\sf \implies \log(x) + \log(y) = \log(xy)$

So,

$\sf \log(x + \sqrt{ {x}^{2} - 1} ) + \log(x - \sqrt{ {x}^{2} - 1 } )$

$\sf = \log \big((x + \sqrt{ {x}^{2} - 1} )(x - \sqrt{ {x}^{2} - 1} ) \big)$

Using identity (a + b)(a - b) = a² - b², we get,

$\sf = \log \big( {(x)}^{2} - (\sqrt{ {x}^{2} - 1} )^{2} \big)$

$\sf = \log \big( {x}^{2} - ({x}^{2} - 1)\big)$

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

$\sf = \log \big( 1\big)$

Since base is not given, we will consider the base 10 logarithm.

Let log(1) = n

➡ 10ⁿ = 1

➡ 10ⁿ = 10⁰ (As any number except 0 raised to the power 0 is always 1)

➡ n = 0

Hence,

log(1) = 0

= RHS (Proved)

Therefore,

$\sf \implies\log(x + \sqrt{ {x}^{2} - 1} ) + \log(x - \sqrt{ {x}^{2} - 1 } ) = 0$

Hence Proved.

Answer 2

$\sf{\log _{10}\left(x+\sqrt{x^2-1}\right)+\log _{10}\left(x-\sqrt{x^2-1}\right)}$

Apply Log Rule : $\bf{\log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)}$

$\sf{=\log _{10}\left(\left(x+\sqrt{x^2-1}\right)\left(x-\sqrt{x^2-1}\right)\right)}$

Apply Difference of Two Squares Formula : $\bf{(a+b)(a-b)=a^2-b^2}$

$\sf{=\log _{10}\left(x^2-\left(\sqrt{x^2-1}\right)^2\right)}$

$\sf{=\log _{10}\left(x^2-\left(x^2-1\right)\right)}$

$\sf{=\log _{10}\left(x^2-x^2+1\right)}$

$\sf{=\log _{10}\left(1\right)}$

Apply Log Rule : $\bf{\log _a\left(1\right)=0}$

$\sf{=0}$

Hence Proved !