# If $x > 0$ and ${\log _2}x + {\log _2}\left( {\sqrt x } \right) + {\log _2}\left( {^4\sqrt x } \right) + \log \left( {^8\sqrt x } \right) + ..................\infty = 4$, then $x$ is

(a) 2

(b) 3

(c) 4

(d) 5

Last updated date: 27th Mar 2023

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Answer

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Hint- Use the property of addition of multiple logarithm terms followed by converting that into a form of a progression to ease the simplification.

Its been given that for $x > 0$,

${\log _2}x + {\log _2}\left( {\sqrt x } \right) + {\log _2}\left( {^4\sqrt x } \right) + \log \left( {^8\sqrt x } \right) + ..................\infty = 4$

Now using the logarithm property that is $\log A + \log B + \log C......... = \log (ABC..............)$

We can write above expression as

${\log _2}(x\left( {\sqrt x } \right)\left( {^4\sqrt x } \right)\left( {^8\sqrt x } \right)..................\infty ) = 4$

Writing in terms of powers, we have

${\log _2}\left[ {x \times {x^{\dfrac{1}{2}}} \times {x^{\dfrac{1}{4}}} \times {x^{\dfrac{1}{8}}}.................\infty } \right] = 4$

Using property of power addition we have

${\log _2}\left[ {{x^{1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ..............\infty }}} \right] = 4$

Now clearly the power of x is forming a Geometric Progression with first term as $1$ and common ratio as $\dfrac{1}{2}$ and also it is an $\infty $ GP.

We know that the sum of an infinite GP is $\dfrac{a}{{1 - r}}$ where $a$ is first term and $r$ is common ratio.

So, we have

${\log _2}{x^{\dfrac{1}{{1 - \left( {\dfrac{1}{2}} \right)}}}} = {\log _2}{x^2} = 4$

Now we have,

${\log _2}{x^2} = 4$

Using the property of logarithm that is ${\log _a}b = p \Rightarrow b = {a^p}$

We get ${x^2} = {2^4} = 16$

$\therefore x = \pm 4$

But since the domain of log is always a number greater than 0 hence $x$ can’t be equal to $-4$, so $x = 4$ is the only right answer.

Hence, option(c) is correct.

Note - In these types of problems, try to simplify the given expression using properties of logarithmic expressions. The simplification is followed by identifying some property of the infinite series i.e. geometric progression in this case and the rest is solved using G.P. formulas.

Its been given that for $x > 0$,

${\log _2}x + {\log _2}\left( {\sqrt x } \right) + {\log _2}\left( {^4\sqrt x } \right) + \log \left( {^8\sqrt x } \right) + ..................\infty = 4$

Now using the logarithm property that is $\log A + \log B + \log C......... = \log (ABC..............)$

We can write above expression as

${\log _2}(x\left( {\sqrt x } \right)\left( {^4\sqrt x } \right)\left( {^8\sqrt x } \right)..................\infty ) = 4$

Writing in terms of powers, we have

${\log _2}\left[ {x \times {x^{\dfrac{1}{2}}} \times {x^{\dfrac{1}{4}}} \times {x^{\dfrac{1}{8}}}.................\infty } \right] = 4$

Using property of power addition we have

${\log _2}\left[ {{x^{1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ..............\infty }}} \right] = 4$

Now clearly the power of x is forming a Geometric Progression with first term as $1$ and common ratio as $\dfrac{1}{2}$ and also it is an $\infty $ GP.

We know that the sum of an infinite GP is $\dfrac{a}{{1 - r}}$ where $a$ is first term and $r$ is common ratio.

So, we have

${\log _2}{x^{\dfrac{1}{{1 - \left( {\dfrac{1}{2}} \right)}}}} = {\log _2}{x^2} = 4$

Now we have,

${\log _2}{x^2} = 4$

Using the property of logarithm that is ${\log _a}b = p \Rightarrow b = {a^p}$

We get ${x^2} = {2^4} = 16$

$\therefore x = \pm 4$

But since the domain of log is always a number greater than 0 hence $x$ can’t be equal to $-4$, so $x = 4$ is the only right answer.

Hence, option(c) is correct.

Note - In these types of problems, try to simplify the given expression using properties of logarithmic expressions. The simplification is followed by identifying some property of the infinite series i.e. geometric progression in this case and the rest is solved using G.P. formulas.

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