1
$\begingroup$

How to solve this exponential question without the log function?

I could just use derivative method (the log approach) to get the numbers of x that satisfy:

$3^x=6x+2$

What is another method to get the number of x?

  • 0
    @Victor For "explicit," Lambert's $W$: one can sidestep mentioning $\log$. But I do not really consider Lambert's $W$ explicit. Unless one considers solutions of algebraic DE explicit, which is quite a stretch of the ordinary meaning.2012-10-05

3 Answers 3

2

We want to find the roots of $f(x)= 3^x-6x-2.$

\begin{align*} f'(x)&=(\ln 3) 3^x - 6 \\ f''(x)&=(\ln 3)^2 3^x >0 \end{align*}

Note that $f(-1)>0, f(0)<0,f(3)>0$. Hence there must be exactly one root each in $(-1,0)$ and $(0,3)$. You can find both roots using standard methods like Newton-Raphson.

1

This is possible to solve analytically if you use Lambert's W function. The Wikipedia page on the topic says that $ p^{ax+b} = cx+d $ has solutions $ x = -\frac{W(-\frac{a \ln p}{c}p^{b-\frac{ad}{c}})}{a\ln p} - \frac{d}{c} $

  • 0
    The connection with the original equation is: if $t = -(x+1/3) \ln 3$, $3^x = 6 x + 2$ is equivalent to $t e^t = $ (a constant that I'll leave you to figure out).2012-10-05
1

Alternative approach:

Since $f(x)=3^x=e^{x\ln3}$ we have $f(x)\geq x\ln 3+1$ (equality only at $0$). It follows that $y=x\ln 3+2$ must intersect $f$ exactly twice. Since $6>\ln 3,\;\;y=6x+2$ must also intersect $f$ exactly twice (i.e. there are exactly two solutions).

(this works only because $f$ is exponential and therefore must exceed any affine function for sufficient large $x$)