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How to solve that equation: $$5^x = -2x + 7$$ I already have the answer $x=1$. Can anyone please explain to me?

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$5^x$ monotonically increases, $-2x+7$ monotonically decreases. So, your equation couldn't have more than one root. And you've found it.

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    $\begingroup$ It is a good solution...but it would be better if it was accompanied with the calculus to support your said statements. $\endgroup$ – Hawk Feb 21 '14 at 14:27
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Since the function $5^x+2x-7$ is continuous and you can see that it is monotonic increasing, by Brouwer's fixed point theorem (along with continuity ) the function has a unique zero which happens to be at $x=1$.

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$x \gt 1 \implies 5^{x} \gt 5^{1} = 5 = -2 \times 1+7 \gt -2x +7$
$x \lt 1 \implies 5^{x} \lt 5^{1} = 5 = -2 \times1+7\lt -2x +7$
Hence only solution $x=1$.

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    $\begingroup$ This verifies that $x = 1$ is the only solution but does not solve the equation. $\endgroup$ – Ishfaaq Feb 21 '14 at 14:28
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    $\begingroup$ This does $solve$ the equation as $a \gt b$ or $a \lt b$ means that $a \neq b$. I have shown that only $1$ satisfies the equation and anything else does not. $\endgroup$ – Indrayudh Roy Feb 21 '14 at 14:31
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    $\begingroup$ What does your statement $5=-2.1+7$ mean? It's wrong, since $-2.1+7=4.9$ $\endgroup$ – Ruslan Feb 21 '14 at 14:45
  • $\begingroup$ Sorry, I meant product, not decimals. $\endgroup$ – Indrayudh Roy Feb 21 '14 at 15:04
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The equation can be solved in terms of the Lambert W function.

Writing $5^x$ as ${\rm e}^{x \ln 5}$ the equation becomes \begin{eqnarray*} {\rm e}^{x \ln 5} &=& -2x + 7 \\ \Rightarrow (-2x + 7) {\rm e}^{-x \ln 5} &=& 1 \\ \frac{\ln 5}{2} (-2x + 7) {\rm e}^{-x \ln 5} &=& \frac{\ln 5}{2} \\ (-x \ln 5 + \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5} &=& \frac{\ln 5}{2} \\ (-x \ln 5 + \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5 + \frac{7}{2} \ln 5 - \frac{7}{2} \ln 5} &=& \frac{\ln 5}{2} \\ (-x \ln 5+ \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5 + \frac{7}{2}\ln 5} {\rm e}^{-\frac{7}{2} \ln 5} &=& \frac{\ln 5}{2} \\ \Rightarrow (-x \ln 5 + \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5 + \frac{7}{2} \ln 5} &=& \frac{\ln 5}{2} {\rm e}^{\frac{7}{2} \ln 5} \end{eqnarray*} The last equation is now in the form for the defining equation for the Lambert W function, namely $${\rm W}(x) {\rm e}^{{\rm W}(x)} = x$$ Solving for W one has $$-x \ln 5 + \frac{7}{2} \ln 5 = {\rm W}_0 \left (\frac{\ln 5}{2} {\rm e}^{\frac{7}{2} \ln 5} \right )$$ Note the principal branch ${\rm W}_0(x)$ for the Lambert W function is chosen since its argument is positive.

Finally, solving for $x$ yields \begin{equation} x = \frac{7}{2} - \frac{1}{\ln 5} {\rm W}_0 \left (\frac{\ln 5}{2} {\rm e}^{\frac{7}{2} \ln 5} \right ) \end{equation} which is the solution we sought.

Getting the above solution into a form more easily recognised we note that since
$${\rm e}^{\frac{7}{2} \ln 5} = {\rm e}^{(\frac{5}{2} + 1) \ln 5} = {\rm e}^{\frac{5}{2} \ln 5} {\rm e}^{\ln 5} = 5 {\rm e}^{\frac{5}{2} \ln 5}$$ the equation for $x$ can be re-written as $$x = \frac{7}{2} - \frac{1}{\ln 5} {\rm W}_0 \left (\frac{5 \ln 5}{2} {\rm e}^{\frac{5 \ln 5}{2}} \right )$$ Now as the Lambert W function is the inverse of the function $y = x {\rm e}^x$, one has the following simplification $${\rm W}_0 \left (\frac{5 \ln 5}{2} {\rm e}^{\frac{5 \ln 5}{2}} \right ) = \frac{5 \ln 5}{2}$$ from which we get $$x = \frac{7}{2} - \frac{1}{\ln 5} \frac{5 \ln 5}{2} = \frac{7}{2} - \frac{5}{2} = 1$$

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