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Can someone explain including working out how to solve this?

$\dfrac{5^x}{x} = 79.85957$

I know that the answer is $x = 3.5$, but how does one normalise the equation so that the x is on one side?

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    Notice the equation $e^x=3$ has no explicit solution in terms of algebraic operations (this excludes the logarithm). However, once we invent the log function then we have an *explicit* solution $\ln 3$. What exactly constitutes an *explicit* solution depends on the universe of allowed functions. Apparently, if we include the Lambert W function then there is an explicit solution. With this understanding the distinction between solving $e^x=3$ and the $5^x/x=79$ is removed. You can argue the natural log is more natural! But, others argue to include Lambert in our lexicon of basic functions.2012-08-22

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See here for a related problem. We can solve the equation in terms of the Lambert W function,

$ \frac{5^x}{x}=c \Rightarrow \frac{ {\rm e}^{x \ln(5)} }{x} = c \Rightarrow \frac{ {\rm e}^{z } }{z} = \frac{c}{\ln(5)}\,, $

where $z=x\ln(5)$. The last equation has the solution $ z= -\operatorname{LambertW} \left( -{\frac {\ln \left( 5 \right) }{c}} \right) \,,$

where the Lambert W function is the solution of the equation $ y{\rm e}^{y}=x \,. $

Substituting $z = x \ln(5)$ and $c=79.85957$ gives the two real solutions

$ x_1 = -\frac{1}{\ln(5)} \operatorname{LambertW}_{-1} \left( -{\frac {\ln \left( 5 \right) }{79.85957}} \right) = 3.499999994 $ and

$ x_2 = -\frac{1}{\ln(5)}\operatorname{LambertW}_{0} \left( -{\frac {\ln \left( 5 \right) }{79.85957}} \right) = 0.01278225404 \,. $

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    @JamesS.Cook: I believe it is a very good idea.2012-08-22