
How do you find the inverse of \[y = {\log _5}x?\]
Answer
531.9k+ views
Hint: whenever we get the problem in which we have to find the inverse of log, we have to follow the following steps so that it can be easily calculated. In the first step if a problem is provided in \[f(x)\] form then we should have changed it to \[y\] . In the second step we are going to interchange the variables and solve for \[y\] . In the last step change \[y\] to \[{f^{ - 1}}(x)\] or \[{y^{ - 1}}\] to get the final required value.
Formula used:
1. If \[{\log _e}x = a\] then it can be written as \[x = {e^a}\]
2. \[{a^{{{\log }_a}x}} = x\]
Complete step by step answer:
First writing above given equation as follows,
\[ \Rightarrow y = {\log _5}x\]
Let \[y\] be equal to \[f(x)\] as it is a function of \[x\] . Therefore, writing it as
\[ \Rightarrow f(x) = {\log _5}x\]
Interchanging \[x\] and \[y\] in the above equation and writing it as,
\[ \Rightarrow x = {\log _5}y\]
Now, raise power of 5 on both sides and write it as following,
\[ \Rightarrow {5^x} = {5^{{{\log }_5}y}}\]
By using above given logs formula \[{5^{{{\log }_5}y}}\] can be written as following,
\[ \Rightarrow {5^{{{\log }_5}y}} = y\]
Now replacing \[{5^x} = {5^{{{\log }_5}y}}\] by above obtained value, we get
\[ \Rightarrow {5^x} = y\]
Now, \[y\] can be replaced by \[{f^{ - 1}}(x)\] as it is solved by interchanging the variables \[x\] and \[y\] .
Therefore, we can write above equation as following,
\[ \Rightarrow {5^x} = {f^{ - 1}}(x)\]
\[{f^{ - 1}}(x)\] is called an inverse function which depends upon \[x\] and can be replaced by \[{y^{ - 1}}\] .
Therefore, we can write above equation as following,
\[ \Rightarrow {5^x} = {y^{ - 1}}\]
It can also be written as,
\[ \Rightarrow {y^{ - 1}} = {5^x}\]
\[{y^{ - 1}}\] is called the inverse of \[y\] . Therefore, we have obtained the required value of the inverse of \[y\] . Which can be written as following,
Inverse of \[y = {5^x}\] .
So, finally we can write Inverse of \[y = {\log _5}x\] as following,
\[ \Rightarrow \] Inverse of \[y = {5^x}\]
Note: In the above given problem, we must have basic knowledge of logs for example what is the base of logs, difference between base 10 and base \[e\] . We should also be familiar with the inverse of function. In the above problem log has base 5.
Formula used:
1. If \[{\log _e}x = a\] then it can be written as \[x = {e^a}\]
2. \[{a^{{{\log }_a}x}} = x\]
Complete step by step answer:
First writing above given equation as follows,
\[ \Rightarrow y = {\log _5}x\]
Let \[y\] be equal to \[f(x)\] as it is a function of \[x\] . Therefore, writing it as
\[ \Rightarrow f(x) = {\log _5}x\]
Interchanging \[x\] and \[y\] in the above equation and writing it as,
\[ \Rightarrow x = {\log _5}y\]
Now, raise power of 5 on both sides and write it as following,
\[ \Rightarrow {5^x} = {5^{{{\log }_5}y}}\]
By using above given logs formula \[{5^{{{\log }_5}y}}\] can be written as following,
\[ \Rightarrow {5^{{{\log }_5}y}} = y\]
Now replacing \[{5^x} = {5^{{{\log }_5}y}}\] by above obtained value, we get
\[ \Rightarrow {5^x} = y\]
Now, \[y\] can be replaced by \[{f^{ - 1}}(x)\] as it is solved by interchanging the variables \[x\] and \[y\] .
Therefore, we can write above equation as following,
\[ \Rightarrow {5^x} = {f^{ - 1}}(x)\]
\[{f^{ - 1}}(x)\] is called an inverse function which depends upon \[x\] and can be replaced by \[{y^{ - 1}}\] .
Therefore, we can write above equation as following,
\[ \Rightarrow {5^x} = {y^{ - 1}}\]
It can also be written as,
\[ \Rightarrow {y^{ - 1}} = {5^x}\]
\[{y^{ - 1}}\] is called the inverse of \[y\] . Therefore, we have obtained the required value of the inverse of \[y\] . Which can be written as following,
Inverse of \[y = {5^x}\] .
So, finally we can write Inverse of \[y = {\log _5}x\] as following,
\[ \Rightarrow \] Inverse of \[y = {5^x}\]
Note: In the above given problem, we must have basic knowledge of logs for example what is the base of logs, difference between base 10 and base \[e\] . We should also be familiar with the inverse of function. In the above problem log has base 5.
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