Question

How do you evaluate ${{e}^{\ln y}}$ ?

Answer 1

To evaluate ${{e}^{\ln y}}$ , let us consider $x={{e}^{\ln y}}$ . We will then have to take the natural logarithm on both sides which yields $\ln x=\ln \left( {{e}^{\ln y}} \right)$ . Now, using the identity$\ln {{x}^{a}}=a\ln x$ , we will get $\ln x=\ln \left( y \right)\ln \left( e \right)$ . When we apply the identity, $\ln e=1$ , we can further simplify the expression to $\ln x=\ln y$ . The last step is to apply the rule, if $\ln a=\ln b$ then $a=b$ , which will result in the required answer.

Complete step by step solution:
We have to evaluate ${{e}^{\ln y}}$ . Let us consider $x={{e}^{\ln y}}$ .
Now, let us take the natural logarithm on both sides. We will get
$\ln x=\ln \left( {{e}^{\ln y}} \right)$
We know that $\ln {{x}^{a}}=a\ln x$ . Hence, the above form can be written as
$\ln x=\ln \left( y \right)\ln \left( e \right)$
We know that the natural logarithm of e is 1, that is, $\ln e=1$ . Hence, the above form becomes
$\ln x=\ln \left( y \right)\times 1=\ln y$
Let us apply the rule, if $\ln a=\ln b$ then $a=b$ . Hence, we can write the above equation as

& \Rightarrow x=y \\\ & \Rightarrow {{e}^{\ln y}}=y \\\ \end{aligned}$$ **Hence the answer is y.** **Note:** Students have a chance of making mistakes when using identities. They may write $\ln {{x}^{a}}=x\ln a$ instead of $\ln {{x}^{a}}=a\ln x$ . Also, they may consider the value of $\ln e$ to be -1 or 0. Note that logarithm of any value is never a negative. We can also solve this problem using the identity ${{a}^{{{\log }_{a}}x}}=x$ . This is explained below. We know that natural logarithm is the logarithm to the base e. This can be expressed as $\ln ={{\log }_{e}}$ . Therefore, we can write ${{e}^{\ln y}}$ as ${{e}^{{{\log }_{e}}y}}$ . Now, when we use the identity ${{a}^{{{\log }_{a}}x}}=x$ here, we will get ${{e}^{{{\log }_{e}}y}}=y$ .