Question

Write the following in logarithmic form: \[\dfrac{1}{{257}} = {4^a}\]

Answer 1

Hint: Logarithm is the inverse function to the exponentiation function. We can take \[log\] to both sides of an equation, it will not change the value of the equation. We can operate logarithm function as per its properties. \[\log {x^n} = n\log x\], \[\dfrac{{\log x}}{{\log y}} = {\log _y}x\]. Here we have first take \[log\] to both sides of the equation then simplify it.

Complete step by step answer:
We have given equation is
\[\dfrac{1}{{257}} = {4^a}\]
We have written the above equation in logarithmic form, it will be more precise to write the above equation in terms of the unknown \[a\]. Applying \[log\] to both the sides of the equation we get,
\[\log \dfrac{1}{{257}} = \log {4^a}\]
As from the properties of logarithm function
\[\log {x^n} = n\log x\]
For this comparison we have\[x = 4\] and \[n = a\].

So applying this property to the left hand side of the equation we have,
\[\log \dfrac{1}{{257}} = a\log 4\]
Now, dividing \[\log 4\] both the sides in order to get the result we get,
\[\dfrac{{\log \dfrac{1}{{257}}}}{{\log 4}} = \dfrac{{a\log 4}}{{\log 4}}\]
Cancelling out \[\log 4\] from left hand side of the equation we get,
\[a = \dfrac{{\log \dfrac{1}{{257}}}}{{\log 4}}\]
Again by the properties of logarithmic function
\[\dfrac{{\log x}}{{\log y}} = {\log _y}x\]
Comparing for \[x\] and \[y\] we observe here \[x = \dfrac{1}{{257}},{\text{ }}y = 4\]
So, applying this property to the Left hand side of the equation we get,
\[\therefore a = {\log _4}\dfrac{1}{{257}}\]

Hence, the logarithmic form of \[\dfrac{1}{{257}} = {4^a}\] is \[a = {\log _4}\dfrac{1}{{257}}\].

Note: The opposite of logarithm is antilogarithm. To remove log from any term of an equation we use antilog on both sides of the equation to cancel it out. Antilog has similar properties as that of the log function. Logarithm of \[1\] is \[\,0\] that is, \[\log 1 = 0\]. In general representation the base of the \[\log \] is either \[10\] or \[e\]. Logarithm functions are applicable for evaluating higher powers of numbers.