Question

How do you solve ${e^{2x}} - \left( {4{e^x}} \right) + 3 = 0$ ?

Answer 1

To solve the given equation, find out the factors of the equation by substituting the exponential term ${e^{2x}}$ as ${\left( {{e^x}} \right)^2}$, hence by this we can get the factors by taking natural logarithm on both the sides of the exponent we can find the value of $x$. As Logarithmic functions are the inverses of exponential functions hence by this, we can get the value of $x$.

Complete step by step solution:
Let us write the given equation
${e^{2x}} - \left( {4{e^x}} \right) + 3 = 0$
Let us rewrite the equation ${e^{2x}}$as ${\left( {{e^x}} \right)^2}$, therefore the equation becomes
${\left( {{e^x}} \right)^2} - \left( {4{e^x}} \right) + 3 = 0$ …………………….. 1
Consider $u = {e^x}$
Substitute $u$ for all occurrences of ${e^x}$, hence the equation 1 is
${u^2} - 4u + 3 = 0$
Now factorize the equation using AC method as it is of the form ${x^2} - bx + c$, in which we need to find pair of integers whose product is c and sum is b, hence
${u^2} - 4u + 3 = 0$
${u^2} - 3u - 1u + 3 = 0$
$\left( {u - 3} \right)\left( {u - 1} \right)$ …………………….. 2
Therefore, $\left( {u - 3} \right)\left( {u - 1} \right)$ is the factored form of the equation.
The integers we got are -3 and -1.
Now replace $u$ with ${e^x}$ in equation 2 we get
$\left( {{e^x} - 3} \right)\left( {{e^x} - 1} \right)$
Therefore, the as per the given equation becomes
$\left( {{e^x} - 3} \right)\left( {{e^x} - 1} \right) = 0$ ……………………. 3
Now let us find each exponential term of equation 3 that is by taking
$\left( {{e^x} - 3} \right) = 0$ and $\left( {{e^x} - 1} \right) = 0$.
For, $\left( {{e^x} - 3} \right) = 0$
Add 3 on both sides of the equation, we get
${e^x} - 3 + 3 = 3$
Therefore,
${e^x} = 3$ …………………………… 4
Take natural logarithm on both the sides of equation 4 to remove the variable from Exponent i.e.,
$\ln \left( {{e^x}} \right) = \ln \left( 3 \right)$
Expand $\ln \left( {{e^x}} \right)$ to move $x$ outside logarithm, hence
$x\ln \left( e \right) = \ln \left( 3 \right)$
As natural log of $e$ is 1
$x\left( 1 \right) = \ln \left( 3 \right)$
$x = \ln \left( 3 \right)$
Now, let us calculate for $\left( {{e^x} - 1} \right) = 0$
Add 1 on both sides of the equation, we get
${e^x} - 1 + 1 = 1$
Therefore,
${e^x} = 1$ …………………………… 5
Take natural logarithm on both the sides of equation 5 to remove the variable from Exponent i.e.,
$\ln \left( {{e^x}} \right) = \ln \left( 1 \right)$
Expand $\ln \left( {{e^x}} \right)$ to move $x$ outside logarithm, hence
$x\ln \left( e \right) = \ln \left( 1 \right)$
As natural log of $e$ is 1
$x\left( 1 \right) = \ln \left( 1 \right)$
$x = \ln \left( 1 \right)$
As the natural logarithm of 1 is 0, therefore the value of $x$ is 0.
Therefore, for the equation ${e^{2x}} - \left( {4{e^x}} \right) + 3 = 0$, the factors obtained is true for
$x = \ln \left( 3 \right),0$
$x = 1.098612$

Note: The key point to find the given equation is that when the equation consists of exponential terms, just take natural logarithm on both the sides of the equation as to solve for the value of $x$ we need to remove the variable from the exponent by taking ln of the function. As Logarithmic functions are the inverses of exponential functions.