# Number to the exponent divided by exponent value

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?

-
This is called a transcendental equation. They are solved by approximations only. – Karolis Juodelė Aug 22 '12 at 10:46
@KarolisJuodelė There's no evidence that each transcendental equation could not be solved out explicitly. And we should point out what is explicit, say, the finite composition of addition, subtraction, multiplying, division and take logarithm, take exponent, and a list of constants (e.g. $1$, $i$, $\pi$, $e$, $\gamma$). – Frank Science Aug 22 '12 at 13:55
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. – James S. Cook Aug 22 '12 at 14:17

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

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

-
maybe I should put a section in my precalculus materials about this... I'm sure the students would appreciate it. – James S. Cook Aug 22 '12 at 14:18
@JamesS.Cook: I believe it is a very good idea. – Mhenni Benghorbal Aug 22 '12 at 21:32