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How to solve $3{e^x} = 2{e^{ - x}} + 4$?

seo-qna
Last updated date: 21st Jul 2024
Total views: 351k
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Answer
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Hint: Here we will rearrange the given equation to get a quadratic equation in ${e^x}$and solve it using quadratic formula to get the value of ${e^x}$then by applying logarithm on both sides we will find the value of $x$. The quadratic formula will give two roots but since we have to apply log we will only consider the positive root.

Complete step-by-step answer:
The given linear equation is $3{e^x} = 2{e^{ - x}} + 4$,
Since the exponential function on the right side has negative power it can be shifted to the denominator as , $3{e^x} = \dfrac{2}{{{e^x}}} + 4$ taking LCM on right side of the equation we get,
$3{e^x} = \dfrac{{2 + 4{e^x}}}{{{e^x}}}$ on cross multiplying we have $3{e^x} \times {e^x} = 2 + 4{e^x}$
Taking all the terms to the left side we get a quadratic equation in ${e^x}$, i.e. $3{({e^x})^2} - 4{e^x} - 2 = 0$
Comparing this with the standard quadratic form $a{x^2} + bx + c = 0$ , in place of $x$ we have ${e^x}$and comparing the coefficients we have $a = 3,b = - 4,c = - 2$.
The quadratic formula for standard form is given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$,substituting for $x,a,b,c$ we get,
$$$$$$\eqalign{
  & {e^x} = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 3 \times ( - 2)} }}{{2 \times 3}} \cr
  & = \dfrac{{4 \pm \sqrt {16 + 24} }}{6} = \dfrac{{4 \pm \sqrt {40} }}{6} = \dfrac{{4 \pm \sqrt {4 \times 10} }}{6} = \dfrac{{4 \pm 2\sqrt {10} }}{6} = \dfrac{{2(2 \pm \sqrt {10} )}}{6} \cr
  & \Rightarrow {e^x} = \dfrac{{2 \pm \sqrt {10} }}{3} \cr} $$
Now $\sqrt {10} > 2$ , hence to get a positive value for ${e^x}$and a real value for $x$ we will consider only the positive sign, i.e. $${e^x} = \dfrac{{2 + \sqrt {10} }}{3}$$
Now taking log to the base $e$ on both sides we get,
$${\log _e}{e^x} = {\log _e}\dfrac{{2 + \sqrt {10} }}{3}$$
By using power rule of logarithm we can write ,
$$x{\log _e}e = {\log _e}\dfrac{{2 + \sqrt {10} }}{3}$$
We know that the value of $${\log _e}e$$is $$1$$.
$$ \Rightarrow x = {\log _e}\dfrac{{2 + \sqrt {10} }}{3}$$
So, the correct answer is “$ x = {\log _e}\dfrac{{2 + \sqrt {10} }}{3}$”.

Note: In questions like this first we need to rearrange the terms to get the variables on one side and if possible try to simplify it. After simplification based on whether we get a linear or quadratic equation we will use respective formulas to find the value of the variable. Linear equation contains variables with degree one and quadratic equation contains variables with degree two.