Question

The logarithm of a number to the base V2 is k. What is its logarithm to the base 2v2 ?​

Answer 1

Answer:

$log_{2\sqrt 2} X = \frac{1}{3}k$

Step-by-step explanation:

Given

Let the number be X

From the first statement, we have:

$log_{\sqrt 2} X = k$

Required

Find $log_{2\sqrt 2} X$

$log_{\sqrt 2} X = k$

using the following law of logarithm

$log_ab = n, b=a^n$

So:

$log_{\sqrt 2} X = k$

$X = \sqrt{2}^k$

Substitute: $X = \sqrt{2}^k$ in $log_{2\sqrt 2} X$

$log_{2\sqrt 2} X = log_{2\sqrt 2} ( \sqrt{2}^k)$

$log_{2\sqrt 2} X = klog_{2\sqrt 2} \sqrt{2}$

Apply the following law:

$log_ab = \frac{log\ b}{log\ a}$

$log_{2\sqrt 2} X = k\frac{log\ \sqrt 2}{log\ {2\sqrt 2}}$

Express the square roots as power

$log_{2\sqrt 2} X = k\frac{log\ 2^\frac{1}{2}}{log\ {2 * 2^\frac{1}{2}}}$

$log_{2\sqrt 2} X = k\frac{log\ 2^\frac{1}{2}}{log\ {2^\frac{3}{2}}}$

using the following law of logarithm

$log_ab = n, b=a^n$

$log_{2\sqrt 2} X = k\frac{\frac{1}{2}log\ 2}{\frac{3}{2}log\ 2}}$

$log_{2\sqrt 2} X = k\frac{\frac{1}{2}}{\frac{3}{2}}}$

Rewrite as:

$log_{2\sqrt 2} X = k * \frac{1}{2} \div\frac{3}{2}$

$log_{2\sqrt 2} X = k * \frac{1}{2} *\frac{2}{3}$

$log_{2\sqrt 2} X = k * \frac{1}{1} *\frac{1}{3}$

$log_{2\sqrt 2} X = \frac{1}{3}k$