Question

solve it, log2^√6+log2^√2/3​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

The value of $log_{2} \sqrt{6} + log_{2} \sqrt{\frac{2}{3} }$ is 1

Step-by-step explanation:

Given : $log_{2} \sqrt{6} + log_{2} \sqrt{\frac{2}{3} }$

To find: Solve the find the value of $log_{2} \sqrt{6} + log_{2} \sqrt{\frac{2}{3} }$

Solution:

Logarithm

A logarithm is a mathematical procedure that establishes the number of times the base, or starting point, must be multiplied by itself to produce the target number.

Given term is $log_{2} \sqrt{6} + log_{2} \sqrt{\frac{2}{3} }$

Taking the log term as the common factor

⇒ $log_{2} \sqrt{6 * \frac{2}{3} }$

⇒ $log_{2} \sqrt{2 * 2}$

⇒ $log_{2} \sqrt{4}$

⇒ $log_{2} 2$

⇒ 1

As we know that the value of $log_{2} 2$ is 1

Therefor the value of $log_{2} \sqrt{6} + log_{2} \sqrt{\frac{2}{3} }$ is 1

Final answer:

The value of $log_{2} \sqrt{6} + log_{2} \sqrt{\frac{2}{3} }$ is 1

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