Question

How do you calculate $ {\log _2}0.5 $ ?

Answer 1

Hint: First we will change the base by using the base rule $ \dfrac{{{{\log }_b}(x)}}{{{{\log }_b}(a)}} = {\log _a}x $ .
Then we will evaluate all the required terms. Then we will apply the property. Here, we are using $ \dfrac{{{{\log }_b}(x)}}{{{{\log }_b}(a)}} = {\log _a}x $ logarithmic property. The value of the logarithmic function $ \ln e $ is $ 1 $ .

Complete step-by-step answer:
 We will first apply the base rule. This rule can be used if $ a $ and $ b $ are greater than $ 0 $ and not equal to $ 1 $ , and $ x $ is greater than $ 0 $ .
 $ \dfrac{{{{\log }_b}(x)}}{{{{\log }_b}(a)}} = {\log _a}x $
Substitute in values for the variables in the change of base formula, by using $ b = 10 $ .
 $ \dfrac{{\log (0.5)}}{{\log (2)}} $
Now we can write $ 0.5 $ as $ \dfrac{1}{2} $ . Hence, the expression can be written as,
\[
   = \dfrac{{\log (0.5)}}{{\log (2)}} \\
   = \dfrac{{\log (\dfrac{1}{2})}}{{\log (2)}} \\
   = \dfrac{{\log ({2^{ - 1}})}}{{\log (2)}} \\
   = \dfrac{{ - \log (2)}}{{\log (2)}} \\
   = - 1 \;
 \]
Hence, the value of $ {\log _2}0.5 $ is $ - 1 $ .
So, the correct answer is “-1”.

Note: Remember the logarithmic property precisely which is $ \dfrac{{{{\log }_b}(x)}}{{{{\log }_b}(a)}} = {\log _a}x $ .
 While comparing the terms, be cautious. After the application of property when you get the final answer, tress back the problem and see if it returns the same values. Evaluate the base and the argument carefully. Also, remember that that $ {\ln _e}e = 1 $