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Question

Question: How do you solve\[{e^{2x}} - 3{e^x} - 4 = 0?\]...

How do you solvee2x3ex4=0?{e^{2x}} - 3{e^x} - 4 = 0?

Explanation

Solution

This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. We need to know the relation between exponent and natural logarithm. Also, we need to know the formulae or conditions involved with natural logarithm and exponential terms. We need to know how to calculate the natural logarithm in the scientific calculator.

Complete step-by-step solution:
The given equation in the question is shown below,
e2x3ex4=0{e^{2x}} - 3{e^x} - 4 = 0
The above equation can also be written as,
{e^{2x}} - 3{e^x} = 4$$$$ \to equation\left( 1 \right)
(When4 - 4is the move from LHS to RHS of the equation it convert into44)
For simplifying the equation(1)\left( 1 \right), we multiply the termln\ln on both sides of the equation(1)\left( 1 \right). So, we get
equation(1)e2x3ex=4equation\left( 1 \right) \to {e^{2x}} - 3{e^x} = 4
lne2x3lnex=ln4equation(2)\ln {e^{2x}} - 3\ln {e^x} = \ln 4 \to equation\left( 2 \right)
We know that,
lnxn=nlnxequation(3)\ln {x^n} = n\ln x \to equation\left( 3 \right)
By using the equation(3)\left( 3 \right) in the equation(2)\left( 2 \right) we get,
2xlne3xlne=ln4equation(4)2x\ln e - 3x\ln e = \ln 4 \to equation\left( 4 \right)
We know that,
ln=1e\ln = \dfrac{1}{e}
So,
lne=1\ln e = 1
By using the above condition in the equation(4)\left( 4 \right)we get,
equation(4)2xlne3xlne=ln4equation\left( 4 \right) \to 2x\ln e - 3x\ln e = \ln 4

2x(1)3x(1)=ln4 2x3x=ln4  2x\left( 1 \right) - 3x\left( 1 \right) = \ln 4 \\\ 2x - 3x = \ln 4 \\\

Here, 2x3x=x2x - 3x = - x
So, we get
x=ln4- x = \ln 4
Let’s multiply-on both sides of the above equation, we get
x=ln4x = - \ln 4
So, the final answer is,
x=ln4x = - \ln 4
By using the scientific calculator we get,
x=ln4=1.3863x = - \ln 4 = - 1.3863

Note: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. We would remember the basic conditions related toln\ln andee. To solve these types of questions we would perform arithmetic operations with terms which have different signs. So, we would remember the following things,
When a negative term is multiplied with a negative term the final answer will be a positive term.
When a positive term is multiplied with a positive term the final answer will be a positive term.
When a negative term is multiplied with a positive term the final answer will be a negative term.