Question

How do you solve \[{{e}^{2x}}-5{{e}^{x}}+6=0\]?

Answer 1

Hint: This question belongs to the topic of pre-calculus. In the question, first we will consider \[{{e}^{x}}\] as t. After that, we will solve the equation according to the quadratic equations using the formula \[t=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. After finding the value of t, we will equate the value of t with \[{{e}^{x}}\]. From here, we will solve the equation, and find the value of x.

Complete step by step answer:
Let us solve this question.
In this question, we have to solve the equation \[{{e}^{2x}}-5{{e}^{x}}+6=0\] or we can say that we have to find the value of x from the equation \[{{e}^{2x}}-5{{e}^{x}}+6=0\].
So, the equation is
\[{{e}^{2x}}-5{{e}^{x}}+6=0\]
The above equation can also be written as
\[\Rightarrow {{\left( {{e}^{x}} \right)}^{2}}-5{{e}^{x}}+6=0\]
Let us consider the term \[{{e}^{x}}\] as t. So, we can write the \[{{e}^{x}}=t\]
The above equation can be written as
\[\Rightarrow {{\left( t \right)}^{2}}-5t+6=0\]
\[\Rightarrow {{t}^{2}}-5t+6=0\]
Now, the above equation is in the form of a quadratic equation. And we know that if quadratic is \[a{{t}^{2}}-bt+c=0\], then the value of t will be \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
In the equation, a=1, b=-5, and c=6.
So, the value of t will be
\[t=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 1\times 6}}{2\times 1}\]
\[\Rightarrow t=\dfrac{5\pm \sqrt{25-24}}{2}=\dfrac{5\pm \sqrt{1}}{2}\]
The above can also be written as
\[\Rightarrow t=\dfrac{5\pm 1}{2}\]
So, the values of t will be \[\dfrac{5+1}{2}\] and \[\dfrac{5-1}{2}\]
Or, we can say that the values of t will be \[\dfrac{6}{2}\] and \[\dfrac{4}{2}\].
Or, we can say that the values of t will be 3 and 2.
So, now the values of t will be equated with \[{{e}^{x}}\].
Hence, we can write
\[{{e}^{x}}=3\] and \[{{e}^{x}}=2\]
As we know that if \[\ln a=b\] can also be written as \[a={{e}^{b}}\]
So, we can write
\[x=\ln 3\] and \[x=\ln 2\].

Therefore, the equation \[{{e}^{2x}}-5{{e}^{x}}+6=0\] satisfies two values of x which are ln3 and ln2.

Note: We should have a better pre-calculus to solve this type of question. We should have knowledge in the topic of quadratic equation and logarithmic functions. We should know that \[{{\left( {{a}^{x}} \right)}^{y}}\] can also be written as \[{{a}^{xy}}\] for solving this type of question. And, also we should know that if \[\ln a=b\], then we can say that \[a={{e}^{b}}\].