Question

If x>0 and ${{\log }_{2}}x+{{\log }_{2}}\sqrt{x}+{{\log }_{2}}\left( \sqrt[4]{x} \right)+{{\log }_{2}}\left( \sqrt[8]{x} \right)+\cdots \cdots +\infty =4$ then x=

& A.2 \\\ & B.3 \\\ & C.4 \\\ & D.5 \\\ \end{aligned}$$

Answer 1

In this question, we need to find the value of x if we are given an equation in terms of logarithmic function. For this, we will first simplify the given equation by using various logarithmic, exponential and arithmetic properties and then calculate the value of x. Properties that we will use are,

& \left( i \right)\sqrt[x]{y}={{\left( y \right)}^{\dfrac{1}{x}}} \\\ & \left( ii \right){{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}\left( m\times n \right) \\\ & \left( iii \right){{x}^{a}}\cdot {{x}^{b}}={{x}^{a+b}} \\\ & \left( iv \right){{\log }_{b}}x=y\Rightarrow x={{b}^{y}} \\\ \end{aligned}$$ (v) Sum of infinite terms in a geometric progression with first term a and common ratio r is given by $S=\dfrac{a}{1-r}$. **Complete step by step answer:** Here we are given equation as, $${{\log }_{2}}x+{{\log }_{2}}\sqrt{x}+{{\log }_{2}}\left( \sqrt[4]{x} \right)+{{\log }_{2}}\left( \sqrt[8]{x} \right)+\cdots \cdots +\infty =4$$. We need to find the value of x. For this let us simplify the given equation. We know that, $\sqrt[x]{y}={{\left( y \right)}^{\dfrac{1}{x}}}$ so writing all terms in the form of exponents we have $${{\log }_{2}}x+{{\log }_{2}}{{\left( x \right)}^{\dfrac{1}{2}}}+{{\log }_{2}}{{\left( x \right)}^{\dfrac{1}{4}}}+{{\log }_{2}}{{\left( x \right)}^{\dfrac{1}{8}}}+\cdots \cdots +\infty =4$$. Now let us use the logarithmic property as below to make the left side of the equation. Simplified ${{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}\left( m\times n \right)$ combining all terms we get, $${{\log }_{2}}\left( x\times {{x}^{\dfrac{1}{2}}}\times {{x}^{\dfrac{1}{4}}}\times {{x}^{\dfrac{1}{8}}}\times \cdots \cdots \infty \right)=4$$. We know by the property of exponents that ${{x}^{a}}\cdot {{x}^{b}}={{x}^{a+b}}$ so using it in the above equation we have, $${{\log }_{2}}\left( {{x}^{1+\dfrac{1}{2}+}}^{\dfrac{1}{4}+\dfrac{1}{8}+\cdots \cdots \infty } \right)=4\cdots \cdots \cdots \left( 1 \right)$$. Here the power of exponent is $$1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots \cdots +\infty $$. Let us solve it. As we can see, this series is in the form of geometric progression. Here first term is 1 and common ratio is $\dfrac{1}{2}\left( \because \dfrac{\dfrac{1}{2}}{1}=\dfrac{1}{2}=\dfrac{\dfrac{1}{4}}{\dfrac{1}{2}} \right)$. We know the sum of infinite terms in a geometric progression with the first term a as and the common ratio as r we have $S=\dfrac{a}{1-r}$. So sum of infinite terms in the series $$1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots \cdots +\infty $$ become $S=\dfrac{1}{1-\dfrac{1}{2}}=\dfrac{1}{\dfrac{1}{2}}=2$. Hence $$1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots \cdots +\infty =2$$. Putting this value in equation (1) we get, ${{\log }_{2}}{{x}^{2}}=4$. Now let us use the logarithmic property given as, ${{\log }_{b}}x=y\Rightarrow x={{b}^{y}}$ so we have ${{x}^{2}}={{2}^{4}}$. Taking square root on both sides we have ${{\left( {{x}^{2}} \right)}^{\dfrac{1}{2}}}={{\left( {{2}^{4}} \right)}^{\dfrac{1}{2}}}$. We know ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ so we get $x={{2}^{2}}\Rightarrow x=4$. Hence the value of x is 4. **So, the correct answer is “Option C”.** **Note:** Students should keep in mind all the properties before solving this sum. Make sure to take the sum of infinite terms in a geometric progression. Students can make mistakes in calculating the common ratio. While applying ${{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}\left( m\times n \right)$ make sure that base is same for both terms.