Question

How do you calculate ${\log _2}4 - {\log _4}2$ ?

Answer 1

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

Complete step-by-step solution:
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 $0$.
Here,
 $
  a = 4 \\
  x = 2 \\
 $
So, now we apply the formula to both the terms.
${\log _a}{x^n} = n{\log _a}x$.
$
   = {\log _2}4 \\
   = {\log _2}{2^2} \\
   = 2{\log _2}2 \\
 $
$
   = {\log _4}2 \\
   = \dfrac{1}{{{{\log }_2}4}} \\
   = \dfrac{1}{{{{\log }_2}{2^2}}} \\
   = \dfrac{1}{{2{{\log }_2}2}} \\
 $
Now we subtract both the terms.
$ = 2{\log _2}2 - \dfrac{1}{{2{{\log }_2}2}}$
But we know that ${\log _a}a = 1$.
Hence, the expression will become,
$
   = 2 - \dfrac{1}{2} \\
   = \dfrac{3}{2} \\
 $
Hence, the value of the expression ${\log _2}4 - {\log _4}2$ is $\dfrac{3}{2}$.


Additional Information: A logarithm is the power to which a number must be raised in order to get some other number. Example: ${\log _a}b$ here, a is the base and b is the argument. Exponent is a symbol written above and to the right of a mathematical expression to indicate the operation of raising to a power. The symbol of the exponential symbol is $e$ and has the value $2.17828$. Remember that $\ln a$ and $\log a$ are two different terms. In $\ln a$ the base is e and in $\log a$ the base is $10$. While rewriting an exponential equation in log form or a log equation in exponential form, it is helpful to remember that the base of exponent.

Note: Remember the logarithmic property precisely which is ${\log _a}{x^n} = n{\log _a}x$.
 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 ${\ln _e}e = 1$