
How do you evaluate \[\log _{64}^{{}}\left( \dfrac{1}{4} \right)\]?
Answer
544.8k+ views
Hint: These types of problems can be solved by considering the solution as any variable. Then consider the equation as equation (1). Now we have to apply the basic logarithm formula and then apply this formula to equation (1). Now, by applying some exponential formulas we will get the solution of the problem.
Complete step by step answer:
From the given question, we are given to solve \[\log _{64}^{{}}\left( \dfrac{1}{4} \right)\].
Let us assume the given solution as N.
\[\log _{64}^{{}}\left( \dfrac{1}{4} \right)=N\].
Let us consider
\[\log _{64}^{{}}\left( \dfrac{1}{4} \right)=N.......\left( 1 \right)\]
As we know that
if \[{{\log }_{a}}N=x;\] then \[{{a}^{x}}=N\]
Now we have to apply the above concept to the equation (1).
By applying the above concept, we get
\[{{64}^{N}}=\dfrac{1}{4}\].
Let us consider this as equation (2).
\[{{64}^{N}}=\dfrac{1}{4}.......\left( 2 \right)\]
Now we have to evaluate the above equation and we have to find the value of N.
We can write \[64\] as \[{{4}^{3}}\] . So, now substitute this in equation (2).
By the formula
\[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\].
Now we have to apply the above concept to the equation (3).
\[{{4}^{3N}}=\dfrac{1}{4}........\left( 4 \right)\]
By the formula
\[\dfrac{1}{a}={{a}^{-1}}\].
By applying the above concept to equation (4).
Let us consider
\[{{4}^{3N}}={{4}^{-1}}........\left( 5 \right)\].
By the formula
\[\begin{align}
& if\text{ }{{a}^{m}}={{a}^{n}} \\
& then\text{ }m=n \\
\end{align}\]
By applying the above concept to the equation (5), we get
\[\begin{align}
& 3N=-1 \\
& N=-\dfrac{1}{3} \\
\end{align}\]
Let us consider this as equation (6).
So, by the equation (6) we obtain the value of N. So it is clear that by solving the equation
\[\log _{64}^{{}}\left( \dfrac{1}{4} \right)\], we get \[N=-\dfrac{1}{3}\] .
Note:
Some students may have a misconception that if \[{{\log }_{a}}N=x;\] then \[{{a}^{x}}=N\] . But we know that if \[{{\log }_{a}}N=x;\] then \[{{a}^{x}}=N\] . If this misconception is followed, then the final answer may get interrupted. So, these misconceptions should be avoided. Also students should avoid calculation mistakes while solving the problem.
Complete step by step answer:
From the given question, we are given to solve \[\log _{64}^{{}}\left( \dfrac{1}{4} \right)\].
Let us assume the given solution as N.
\[\log _{64}^{{}}\left( \dfrac{1}{4} \right)=N\].
Let us consider
\[\log _{64}^{{}}\left( \dfrac{1}{4} \right)=N.......\left( 1 \right)\]
As we know that
if \[{{\log }_{a}}N=x;\] then \[{{a}^{x}}=N\]
Now we have to apply the above concept to the equation (1).
By applying the above concept, we get
\[{{64}^{N}}=\dfrac{1}{4}\].
Let us consider this as equation (2).
\[{{64}^{N}}=\dfrac{1}{4}.......\left( 2 \right)\]
Now we have to evaluate the above equation and we have to find the value of N.
We can write \[64\] as \[{{4}^{3}}\] . So, now substitute this in equation (2).
By the formula
\[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\].
Now we have to apply the above concept to the equation (3).
\[{{4}^{3N}}=\dfrac{1}{4}........\left( 4 \right)\]
By the formula
\[\dfrac{1}{a}={{a}^{-1}}\].
By applying the above concept to equation (4).
Let us consider
\[{{4}^{3N}}={{4}^{-1}}........\left( 5 \right)\].
By the formula
\[\begin{align}
& if\text{ }{{a}^{m}}={{a}^{n}} \\
& then\text{ }m=n \\
\end{align}\]
By applying the above concept to the equation (5), we get
\[\begin{align}
& 3N=-1 \\
& N=-\dfrac{1}{3} \\
\end{align}\]
Let us consider this as equation (6).
So, by the equation (6) we obtain the value of N. So it is clear that by solving the equation
\[\log _{64}^{{}}\left( \dfrac{1}{4} \right)\], we get \[N=-\dfrac{1}{3}\] .
Note:
Some students may have a misconception that if \[{{\log }_{a}}N=x;\] then \[{{a}^{x}}=N\] . But we know that if \[{{\log }_{a}}N=x;\] then \[{{a}^{x}}=N\] . If this misconception is followed, then the final answer may get interrupted. So, these misconceptions should be avoided. Also students should avoid calculation mistakes while solving the problem.
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