Solveeit Logo

Question

Question: How do you write \(y=\ln \left( x-2 \right)\) in exponential form?...

How do you write y=ln(x2)y=\ln \left( x-2 \right) in exponential form?

Explanation

Solution

In this question, we have to find the exponential form of the given logarithmic equation. Thus, we will use the exponential formula and the mathematical rule to get the solution. First, we will put the exponential function on both sides in the equation. Then, we will apply the log-exponential formula eloga=a{{e}^{\log a}}=a on the right hand side of the equation. In the last, we will add 2 on both sides of the equation and make the necessary calculations, to get the required solution for the problem.

Complete step by step solution:
According to the problem, we have to find the exponential form of the given logarithmic equation.
The logarithmic equation given to us is y=ln(x2)y=\ln \left( x-2 \right) -------- (1)
First, we will put the exponential function on both sides in the equation (1), we get
exp(y)=exp(ln(x2))\Rightarrow \exp \left( y \right)=\exp \left( \ln \left( x-2 \right) \right)
Therefore, we get
ey=eln(x2)\Rightarrow {{\operatorname{e}}^{y}}={{\operatorname{e}}^{\ln \left( x-2 \right)}}
Now, we will apply the log-exponential formula eloga=a{{e}^{\log a}}=a on the right hand side of the above equation.
ey=x2\Rightarrow {{\operatorname{e}}^{y}}=x-2
Now, we will add 2 on both sides in the above equation, we get
ey+2=x2+2\Rightarrow {{\operatorname{e}}^{y}}+2=x-2+2
As we know, the same terms with opposite signs cancel out each other, thus we get
ey+2=x\Rightarrow {{\operatorname{e}}^{y}}+2=x which is answer
Therefore, for the equation y=ln(x2)y=\ln \left( x-2 \right) , its exponential form of equation is equal to ey+2=x{{\operatorname{e}}^{y}}+2=x.

Note: While solving this problem, do mention all the steps properly to avoid confusion and mathematical error. One of the alternative methods to solve this problem is take the antilog on both sides of the equation and make the necessary calculations, to get the accurate answer.