
How do you solve $2{{e}^{x}}-1=0$ ?
Answer
543.3k+ views
Hint: In this question, we have to find the value of x. Thus, we will use the logarithm function and the basic mathematical rule to get the solution. We will first add 1 on both sides of the equation, and then divide 2 on both sides. After the necessary calculations, we will put log function on both sides of the equation and then apply the logarithm rule $\log \left( {{a}^{b}} \right)=b\log a$ on the left side of the equation. After that, we will again apply the logarithm rule $\log e=1$ on the left hand side of the equation. After further solving, we will get the required value of x.
Complete step by step solution:
According to the problem, we have to find the value of x.
Thus, we will use the logarithm function and the mathematical rule to get the solution.
The equation give to us is $2{{e}^{x}}-1=0$ ------- (1)
First, we will add 1 on both sides in the above equation, we get
$\Rightarrow 2{{e}^{x}}-1+1=0+1$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow 2{{e}^{x}}=1$
Now, we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2{{e}^{x}}}{2}=\dfrac{1}{2}$
As we know, the same terms in the division cancel out with the quotient 1, we get
$\Rightarrow {{e}^{x}}=\dfrac{1}{2}$
Now, we will put log function on both sides in the above equation, we get
$\Rightarrow \log \left( {{e}^{x}} \right)=\log \left( \dfrac{1}{2} \right)$
Now, we will apply the logarithm function rule $\log \left( {{a}^{b}} \right)=b\log a$ on the left-hand side in the above equation, we get
$\Rightarrow x\log \left( e \right)=\log \left( \dfrac{1}{2} \right)$
Now, we will again apply the logarithm function rule $\log e=1$ on the left-hand side of the above equation, we get
$\Rightarrow x.1=\log \left( \dfrac{1}{2} \right)$
Therefore, we get
$\Rightarrow x=\log \left( \dfrac{1}{2} \right)$
Therefore, for the equation $2{{e}^{x}}-1=0$ , the value of x is equal to $\log \left( \dfrac{1}{2} \right)$.
Note: While solving this problem, do mention all the steps properly to avoid errors and confusion. One of the alternative methods to solve this problem, put the log function on both sides of the equation and then solve the equation, to get the required solution for the problem.
Complete step by step solution:
According to the problem, we have to find the value of x.
Thus, we will use the logarithm function and the mathematical rule to get the solution.
The equation give to us is $2{{e}^{x}}-1=0$ ------- (1)
First, we will add 1 on both sides in the above equation, we get
$\Rightarrow 2{{e}^{x}}-1+1=0+1$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow 2{{e}^{x}}=1$
Now, we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2{{e}^{x}}}{2}=\dfrac{1}{2}$
As we know, the same terms in the division cancel out with the quotient 1, we get
$\Rightarrow {{e}^{x}}=\dfrac{1}{2}$
Now, we will put log function on both sides in the above equation, we get
$\Rightarrow \log \left( {{e}^{x}} \right)=\log \left( \dfrac{1}{2} \right)$
Now, we will apply the logarithm function rule $\log \left( {{a}^{b}} \right)=b\log a$ on the left-hand side in the above equation, we get
$\Rightarrow x\log \left( e \right)=\log \left( \dfrac{1}{2} \right)$
Now, we will again apply the logarithm function rule $\log e=1$ on the left-hand side of the above equation, we get
$\Rightarrow x.1=\log \left( \dfrac{1}{2} \right)$
Therefore, we get
$\Rightarrow x=\log \left( \dfrac{1}{2} \right)$
Therefore, for the equation $2{{e}^{x}}-1=0$ , the value of x is equal to $\log \left( \dfrac{1}{2} \right)$.
Note: While solving this problem, do mention all the steps properly to avoid errors and confusion. One of the alternative methods to solve this problem, put the log function on both sides of the equation and then solve the equation, to get the required solution for the problem.
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