In this logarithmic problem, a quantity is expressed in logarithmic form. In this log term, the base of logarithm is square root of $2$ and the quantity is defined in the form of square roots of $2$ as follows.

$\log_{\sqrt{2}}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}}$

The square root of $2$ is the base In this logarithmic expression. For evaluating the given log expression, the square root of $2$ can be written in exponential notation as $2$ raised to the power of quotient of $1$ by $2$.

$=\,\,\,$ $\log_{2^{\frac{1}{2}}}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}}$

The exponent $1$ over $2$ at the base base position of the logarithmic expression can be taken out as a multiplying factor by the base power rule of logarithms.

$=\,\,\,$ $\dfrac{1}{\Big(\dfrac{1}{2}\Big)} \times \log_{2}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}}$

Now, let us simplify the factors in this logarithmic expression mathematically.

$=\,\,\,$ $\dfrac{1}{\dfrac{1}{2}} \times \log_{2}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}}$

$=\,\,\,$ $1 \times \dfrac{2}{1} \times \log_{2}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}}$

$=\,\,\,$ $1 \times 2 \times \log_{2}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}}$

$=\,\,\,$ $2 \times \log_{2}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}}$

$=\,\,\,$ $2\log_{2}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}}$

In this logarithmic problem, the quantity is defined in the form of square roots and number $2$. In fact, we are confused initially to find the values for these types of quantities. However, we can calculate them by expressing them in exponential form.

$=\,\,\,$ $2\log_{2}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}}$

Now, let’s express each square root of $2$ in exponential notation for simplifying the quantity.

$=\,\,\,$ $2\log_{2}{\Bigg(2\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}\Bigg)^{\Large \frac{1}{2}}}$

$=\,\,\,$ $2\log_{2}{\Bigg(2 \times \sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}\Bigg)^{\Large \frac{1}{2}}}$

The power of a product rule can be used to simplify the quantity in the logarithmic expression.

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times \Bigg(\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}\Bigg)^{\Large \frac{1}{2}}\Bigg)}$

Once again, express the square root of 2 in exponential notation for the second factor of the quantity.

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times \Bigg(\Bigg(2\sqrt{2\sqrt{2\sqrt{2}}}\Bigg)^{\Large \frac{1}{2}}\Bigg)^{\Large \frac{1}{2}}\Bigg)}$

The power of an exponential form quantity can be simplified by the power rule of exponents.

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times \Bigg(2\sqrt{2\sqrt{2\sqrt{2}}}\Bigg)^{\Large \frac{1}{2} \large \times \Large \frac{1}{2}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times \Bigg(2\sqrt{2\sqrt{2\sqrt{2}}}\Bigg)^{\Large \frac{1 \times 1}{2 \times 2}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times \Bigg(2\sqrt{2\sqrt{2\sqrt{2}}}\Bigg)^{\Large \frac{1}{4}}\Bigg)}$

Repeat the same procedure for eliminating the square roots completely from the quantity.

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times \Bigg(2 \times \sqrt{2\sqrt{2\sqrt{2}}}\Bigg)^{\Large \frac{1}{4}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times \Bigg(\sqrt{2\sqrt{2\sqrt{2}}}\Bigg)^{\Large \frac{1}{4}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times \Bigg(\Big(2\sqrt{2\sqrt{2}}\Big)^{\Large \frac{1}{2}}\Bigg)^{\Large \frac{1}{4}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times \Big(2\sqrt{2\sqrt{2}}\Big)^{\Large \frac{1}{2} \large \times \Large \frac{1}{4}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times \Big(2\sqrt{2\sqrt{2}}\Big)^{\Large \frac{1 \times 1}{2 \times 4}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times \Big(2\sqrt{2\sqrt{2}}\Big)^{\Large \frac{1}{8}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times \Big(2 \times \sqrt{2\sqrt{2}}\Big)^{\Large \frac{1}{8}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times 2^{\Large \frac{1}{8}} \times \Big(\sqrt{2\sqrt{2}}\Big)^{\Large \frac{1}{8}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times 2^{\Large \frac{1}{8}} \times \Big((2\sqrt{2})^{\Large \frac{1}{2}}\Big)^{\Large \frac{1}{8}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times 2^{\Large \frac{1}{8}} \times (2\sqrt{2})^{\Large \frac{1}{2} \large \times \Large \frac{1}{8}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times 2^{\Large \frac{1}{8}} \times (2\sqrt{2})^{\Large \frac{1 \times 1}{2 \times 8}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times 2^{\Large \frac{1}{8}} \times (2\sqrt{2})^{\Large \frac{1}{16}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times 2^{\Large \frac{1}{8}} \times 2^{\Large \frac{1}{16}} \times (\sqrt{2})^{\Large \frac{1}{16}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times 2^{\Large \frac{1}{8}} \times 2^{\Large \frac{1}{16}} \times \Big(2^{\Large \frac{1}{2}}\Big)^{\Large \frac{1}{16}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times 2^{\Large \frac{1}{8}} \times 2^{\Large \frac{1}{16}} \times 2^{\Large \frac{1}{2} \large \times \Large \frac{1}{16}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times 2^{\Large \frac{1}{8}} \times 2^{\Large \frac{1}{16}} \times 2^{\Large \frac{1 \times 1}{2 \times 16}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2}} \times 2^{\Large \frac{1}{4}} \times 2^{\Large \frac{1}{8}} \times 2^{\Large \frac{1}{16}} \times 2^{\Large \frac{1}{32}}\Bigg)}$

Finally, we have successfully eliminated the square root form by expressing the quantity in the product form of the exponential form quantities.

The bases of exponential form quantities in the product are the same. Hence, the product of them can be simplified by the product rule of exponents with the same base.

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1}{2} \large + \Large \frac{1}{4}\large + \Large \frac{1}{8}\large + \Large \frac{1}{16}\large + \Large \frac{1}{32}}\Bigg)}$

In the exponent position, the quantities are in sum form but they are unlike fractions. However, the sum of them can be evaluated by the addition of unlike fractions.

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{1\times 16 + 1 \times 8 + 1 \times 4 + 1 \times 2 + 1 \times 1}{32}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{16+8+4+2+1}{32}}\Bigg)}$

$=\,\,\,$ $2\log_{2}{\Bigg(2^{\Large \frac{31}{32}}\Bigg)}$

The logarithm of an exponential form quantity can be simplified by the power rule of logarithms.

$=\,\,\,$ $2 \times \dfrac{31}{32} \times \log_{2}{(2)}$

It can be simplified by the log of base rule.

$=\,\,\,$ $2 \times \dfrac{31}{32} \times 1$

$=\,\,\,$ $\dfrac{2 \times 31 \times 1}{32}$

$=\,\,\, \require{cancel}$ $\dfrac{\cancel{2} \times 31 \times 1}{\cancel{32}}$

$=\,\,\,$ $\dfrac{1 \times 31 \times 1}{16}$

$=\,\,\,$ $\dfrac{31 \times 1}{16}$

$=\,\,\,$ $\dfrac{31}{16}$

Latest Math Topics

Latest Math Problems

Email subscription

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.