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# How do you solve for x in $2{e^{x - 2}} = {e^x} + 7$ ?

Last updated date: 04th Aug 2024
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Hint:In this question, an algebraic expression containing one unknown variable quantity is given to us. We know that we need an “n” number of equations to find the value of “n” unknown variables. We have 1 unknown quantity and exactly one equation to find the value of x. We are given an exponential function, where Euler’s number (e) is raised to some power on both the left and the right sides. So, for finding the value of x, we will use the knowledge of exponential functions and rearrange the equation such that x lies on one side of the equation and all other terms lie on the other side. Then by applying the given arithmetic operations, we can find the value of x.

We are given that $2{e^{x - 2}} = {e^x} + 7$
Taking ${e^x}$ to the left-hand side –
$2\dfrac{{{e^x}}}{{{e^2}}} - {e^x} = 7$
Taking ${e^x}$ common –
${e^x}(2{e^{ - 2}} - 1) = 7$
Now, we will take $2{e^{ - 2}} - 1$ to the right-hand side so that all the terms containing x remain on one side and the other side contains constant values –
${e^x} = \dfrac{7}{{2{e^{ - 2}} - 1}}$
Now, we know that
${e^x} = a \\ \Rightarrow x = \ln a \\$
So, we get –
$x = \ln (\dfrac{7}{{2{e^{ - 2}} - 1}})$
We also know that $\log \dfrac{a}{b} = \log a - \log b$
So, we get –
$x = \ln 7 - \ln (2{e^{ - 2}} - 1)$
Hence, when $2{e^{x - 2}} = {e^x} + 7$ , we get $x = \ln (\dfrac{7}{{2{e^{ - 2}} - 1}})$ or $x = \ln 7 - \ln(2{e^{ - 2}} - 1)$ .

Note: In the above solution have used one law of exponents that states that when two exponential numbers having the same base are in division then it is equal to that base raised to the power the difference of the two powers and vice-versa, that is, $\dfrac{{{a^b}}}{{{a^c}}} = {a^{b - c}}\,or\,{a^{b - c}} = \dfrac{{{a^b}}}{{{a^c}}}$ . We have also used the knowledge of logarithm functions in this question, known as the natural logarithm.