Question

If
${a}^{4 } + {b}^{4} = 14 {a}^{2} {b}^{2}$
Then Show That
$log_{e}( {a}^{2} + {b}^{2} ) \: \: \: = \: \: \: log_{e}a \: + log_{e}b \: + 2 log_{e}2$
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Answer 1

Step-by-step explanation:

${a}^{4 } + {b}^{4} = 14 {a}^{2} {b}^{2} \\ = > ({ {a}^{2} })^{2} + ( { {b}^{2} })^{2} = 14 {a}^{2} {b}^{2} \\ = > { ( {a}^{2} + {b}^{2} )}^{2} - 2 {a}^{2} {b}^{2} = 14 {a}^{2} {b}^{2} \\ = > { ( {a}^{2} + {b}^{2} )}^{2} = 14 {a}^{2} {b}^{2} + 2 {a}^{2} {b}^{2} \\ = > { ( {a}^{2} + {b}^{2} )}^{2} = 16 {a}^{2} {b}^{2} \\ = > ( {a}^{2} + {b}^{2} ) = \sqrt{16 {a}^{2} {b}^{2} } \\ = > ( {a}^{2} + {b}^{2} ) = 4ab$

Now,putting the above value,$lhs ,\: log_{e}( {a}^{2} + {b}^{2} ) \\ = log_{e}(4ab) \\ = log_{e}(4) + log_{e}( a) + log_{e}(b) \\ = log_{e}(a) \: + log_{e}(b) \: + log_{e}( {2}^{2} ) \\ = log_{e}(a) \: + log_{e}(b) \: + 2log_{e}( 2) \\ = rhs$

Hence proved!!

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Used formula :

●(x+y)²=x²+y²+2xy => x²+y²=(x+y)²-2xy

$●log(xy) = log(x) + log(y) \\ ● log( {x}^{y} ) = y log(x)$

Answer 2

Formulas used :

$\star \: ( {x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy$

$\star \: log(xy) = log(x) + log(y)$

$\star \: log( {x}^{y} ) = ylog(x)$