Question

Evaluate :- y = log( base {2+√3} ) [ √(7 - √48)] .​

Answer 1

Answer:

-y = {sqrt(7 - 4 sqrt(3)) log(3.732 bp (base pairs))}

Step-by-step explanation:

Answer 2

Answer:

→ y = -1.

Step-by-step explanation:

$\sf \because y = log_{2 + \sqrt{3} }( \sqrt{7 - \sqrt{48} } ) .$

Now, we have

$\sf \because \sqrt{7 - \sqrt{48} } . \\ \\ = \sf \sqrt{7 - \sqrt{16 \times 3} }. \\ \\ = \sf \sqrt{7 - 4 \sqrt{3} } . \\ \\ = \sf \sqrt{ {2}^{2} + {( \sqrt{3} )}^{2} - 2(2)( \sqrt{3} )} . \\ \\ \sf = \sqrt{ {(2 - \sqrt{3}) }^{2} } \: \: or \: \: \sqrt{ {( \sqrt{3} - 2) }^{2} } . \\ \\ \sf = |2 - \sqrt{3} | \: \: or \: \: | \sqrt{3} - 2 | . \\ \\ = \sf2 - \sqrt{3} \: \: or \: \: - ( \sqrt{3 } - 2). \\ \\ \sf = (2 - \sqrt{3} ).$

Now,

$\sf \because y = log_{2 + \sqrt{3} }( \sqrt{7 - \sqrt{48} } ) . \\ \\ \sf \implies y = log_{2 + \sqrt{3} }( 2 - \sqrt{3} ) . \\ \\ \implies \sf(2 - \sqrt{3} ) = {(2 + \sqrt{3} )}^{y} . \\ \\ \tiny \bigg( \tt \because(2 + \sqrt{3} )(2 - \sqrt{3} ) = 1. \implies(2 - \sqrt{3} ) = {(2 + \sqrt{3} )}^{ - 1} . \bigg) \\ \\ \sf \implies {(2 + \sqrt{3}) }^{ - 1} = {(2 + \sqrt{3}) }^{y} . \\ \\ \boxed{ \therefore \pink{ \it y = - 1.}}$

Hence, it is solved.