Question

Hii guys, what's up, .pls solve this question 30,27,26 only

Answer 1

Hiii. .....friend

The answer is here,


26) $= > \: {x}^{2} + {y}^{2} = 23xy$
Adding 2xy on both sides.

$= > \: {x}^{2} + {y}^{2} + 2xy = 23xy + 2xy$

$= > \: ( {x + y)}^{2} = 25xy$

$= > \: x + y = \sqrt{25xy} = 5 \sqrt{xy}$

$= > \: \frac{x + y}{5} = \sqrt{xy}$

Join log on both sides.

$= > \: log( \frac{x + y}{5} ) = log( \sqrt{xy} )$

$= > \: log( \frac{x + y}{5} ) = log( {xy)}^{ \frac{1}{2} }$

$= > \: log( \frac{x + y}{5} ) = \frac{1}{2}( log(x) + log(y) )$
Hence, Proved.

$27) \: p = log_{10}(20) \: and \: \: q = log_{10}(25)$

Given that,

$= > \: 2 log_{10}(x + 1) = 2p - q$

$= > \: 2 log_{10}(x + 1) = 2 log_{10}(20) - log_{10}(25)$

$= > \: log_{10}( {x + 1)}^{2} = log_{10}( \frac{400}{25} )$
If we compare LHS and RHS . We get,

$= > ({x + 1)}^{2} = \frac{400}{25}$
$= > \: x + 1 = \sqrt{ \frac{400}{25} } = \frac{20}{5}$

$= > \: x + 1 = 4$

$= > \: x = 3$

$30) \: log_{a}(x) = \frac{1}{ \alpha } = \frac{ log(x) }{ log(a) }$

$= > \: log(a) = \alpha log(x)$

$log_{b}(x) = \frac{1}{ \beta } = \frac{ log(x) }{ log(b) }$

$= > \: log(b) = \beta log(x)$

$log_{c}(x) = \frac{1}{ \gamma } = \frac{ log(x) }{ log(c) }$
$= > \: log(c) = \gamma log(x)$

$= > \: log_{abc}(x) = \frac{ log(x) }{ log(a) + log(b) + log(c) }$
$= > \: \frac{ log(x) }{ \alpha log(x) + \beta log(x) + \gamma log(x) }$

$= > \: \frac{ log(x) }{ log(x) ( \alpha + \beta + \gamma )}$

$= > \: \frac{1}{ \alpha + \beta + \gamma }$

:-)Hope it helps u.