Question

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Answer 1

Step-by-step explanation:

log10(-1)^1/3= 0

because log10=1

and log1=0

so ,we can say that

log10(-1)=0 answer

option (A) is correct

I think so

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\red{\rm :\longmapsto\: log \bigg(\bigg[2 + \sqrt{5} \bigg]^{\dfrac{1}{3} } +\bigg[2 - \sqrt{5} \bigg]^{\dfrac{1}{3} } \bigg) }$

To evaluate this value, Let first evaluate

$\blue{\bf :\longmapsto\:\bigg[2 + \sqrt{5} \bigg]^{\dfrac{1}{3} } + \bigg[2 - \sqrt{5} \bigg]^{\dfrac{1}{3} }}$

Let assume that,

${\bf :\longmapsto\:x \: = \: \bigg[2 + \sqrt{5} \bigg]^{\dfrac{1}{3} } + \bigg[2 - \sqrt{5} \bigg]^{\dfrac{1}{3} }}$

On Cubing both sides, we get

$\rm :\longmapsto\: {x}^{3} = [2 + \sqrt{5} ] + [2 - \sqrt{5} ] + 3[2 + \sqrt{5} ][2 - \sqrt{5} ]\bigg(\bigg[2 + \sqrt{5} \bigg]^{\dfrac{1}{3} } + \bigg[2 - \sqrt{5} \bigg]^{\dfrac{1}{3} }\bigg)$

$\red{\bigg \{ \because \tt \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \bigg \}}$

Also, we know that,

$\red{\boxed{ \tt{ (x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

$\rm :\longmapsto\: {x}^{3} = 4 + 3(4 - 5)(x)$

$\rm :\longmapsto\: {x}^{3} = 4 + 3( - 1)(x)$

$\rm :\longmapsto\: {x}^{3} = 4 - 3x$

$\rm :\longmapsto\: {x}^{3} + 3x - 4 = 0$

can be rewritten as

$\rm :\longmapsto\: {x}^{3} - x + 4x - 4 = 0$

$\rm :\longmapsto\:x( {x}^{2} - 1) + 4(x - 1) = 0$

$\rm :\longmapsto\:x( {x} - 1)(x + 1)+ 4(x - 1) = 0$

$\rm :\longmapsto\:(x - 1)( {x}^{2} + x + 4) = 0$

$\bf\implies \:x = 1$

and

$\rm :\longmapsto\: {x}^{2} + x + 4 = 0 \: has \: no \: real \: root \: as \: {b}^{2} - 4ac < 0$

Hence,

The value of

${\bf :\longmapsto\: \: \bigg[2 + \sqrt{5} \bigg]^{\dfrac{1}{3} } + \bigg[2 - \sqrt{5} \bigg]^{\dfrac{1}{3} }} = 1$

Therefore,

${\bf :\longmapsto\: \: log\bigg(\bigg[2 + \sqrt{5} \bigg]^{\dfrac{1}{3} } + \bigg[2 - \sqrt{5} \bigg]^{\dfrac{1}{3} }} \bigg)$

$\rm \: = \: \: log(1)$

$\rm \: = \: \: 0$

Hence,

The value of

${\bf :\longmapsto\: \: log\bigg(\bigg[2 + \sqrt{5} \bigg]^{\dfrac{1}{3} } + \bigg[2 - \sqrt{5} \bigg]^{\dfrac{1}{3} }} \bigg) = 0$

Thus,

• Option (A) is correct.

Additional Information :-

$\boxed{ \rm{ log(xy) = log(x) + log(y) }}$

$\boxed{ \rm{ log( \frac{x}{y} ) = log(x) - log(y) }}$

$\boxed{ \rm{ log( {x}^{y} ) = y log(x)}}$

$\boxed{ \rm{ log_{x}(y) = \frac{logy}{logx}}}$

$\boxed{ \rm{ log_{x}(x) = 1}}$

$\boxed{ \rm{ {e}^{logx} = x}}$

$\boxed{ \rm{ {e}^{ylogx} = {x}^{y} }}$

$\boxed{ \rm{ {a}^{y log_{a}(x) } = {x}^{y} }}$

$\boxed{ \rm{ {a}^{ log_{a}(x) } = x }}$