Question

How do you write ${{b}^{y}}=x$ into logarithmic form?

Answer 1

Hint: In this question, we have to write the given function in terms of logarithmic form. As we know, the logarithmic function is the inverse of exponential function. We will apply the logarithmic function when a number has some exponent, thus we will apply the logarithm function on that term and thus convert the exponent as a number. Thus, in this problem, we will put the log function on both sides and then apply the logarithmic function $\log {{a}^{b}}=b\log a$ . After the necessary calculations, we will again apply the logarithmic function $\dfrac{{{\log }_{c}}a}{{{\log }_{c}}b}={{\log }_{b}}a$ in the equation, to get the required solution for the problem.

Complete step by step solution:
According to the question, we have to simplify the given algebraic term in terms of logarithmic function .
Thus, we will use the logarithmic function and the basic mathematical rules to get the solution.
The term given to us is ${{b}^{y}}=x$ -------------- (1)
Now, we will first put the log function on both sides in the equation (1), we get
$\Rightarrow \log \left( {{b}^{y}} \right)=\log x$
Now, we will apply the logarithm function formula $\log {{a}^{b}}=b\log a$ in the above equation, we get
$\Rightarrow y\log b=\log x$
Now, we will divide log b on both sides in the above equation, we get
$\Rightarrow \dfrac{y\log b}{\log b}=\dfrac{\log x}{\log b}$
As we know, same terms in the division cancel out to 1, thus we get
$\Rightarrow y=\dfrac{\log x}{\log b}$
Now, we will again apply the logarithmic function $\dfrac{{{\log }_{c}}a}{{{\log }_{c}}b}={{\log }_{b}}a$ in the above equation, we get
$\Rightarrow y={{\log }_{b}}x$ which is the required solution.
Therefore, for the given term ${{b}^{y}}=x$ , its simplified value in terms of logarithmic function is $y={{\log }_{b}}x$ .

Note: While solving this problem, do the step-by-step calculations properly to avoid confusion and mathematical error. Do not confuse that $\dfrac{\log a}{\log b}$ is equal to ${{\log }_{b}}a$and not $\log \left( b-a \right)$ . Always remember that the logarithm function is the inverse of exponential function, thus we can convert an exponential into logarithm and vice-versa.