0

I'm working through some logs and exponentials questions at the moment in order so that I might be a little prepared for any I might utilize in a science PHD. I'm currently getting through the questions and learning the concepts and am satisfied I'm learning.

Right now I'm looking at some questions which are asking me to derive the value of $x$ to $3$ decimal places. For $3^x = 10$, for example, I deduced $x$ to be $2.096$ having gone through the laws of logarithms.

I have now got to a question that asks me to find the value of $x$ to $3$.d.p and this is it:

$4^x = 5x^{-1}$

I'm just a bit stumped here, I know I should do the same to both sides of the equation, but should I use log rules, or edit my equation in the first place?

Thank you!

M47145
  • 4,106
New Zealand's finest
  • 353

2 Answers2

5

Since we have $4^x=5/x$, I would suggest that we manipulate the equation to be $x\cdot4^x=5$.

The reason for this particular expression is because it resembles the Lambert W-Function

Doing so will lead to the solution: $x=\dfrac{W(\ln(2))}{2\ln(2)}$ where $W(z)$ is the Lambert function. While I am not quite sure what the motivation for an approximation would be, the expression approximates to $1.095$

John
  • 405
1

Just as Leanne answered, there is an explicit solution in terms of Lambert function. Looking at the examples given in the link, you would find $$x=\frac{W(5 \log (4))}{\log (4)}\approx 1.0953$$ More generally, $$a^x=\frac b x \implies x=\frac{W(b \log (a))}{\log (a)}$$

The link also gives series expansions for $W(z)$ (for small and large values of argument $z$).

If you do not want (or cannot) use Lambert function, only numerical methods will give the solution (Newton could be the simplest).

Claude Leibovici
  • 260,315