Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:


What is another method to get the number of x?

share|cite|improve this question
What do you mean here by 'the derivative method'? – sourisse Oct 5 '12 at 21:28
It's going to be hard analytically with an $x$ in both the index of one term and the base of another. – Daryl Oct 5 '12 at 21:32
@Daryl - I think the charm of math is the will of people to increase the difficulty of the solution by using other method. – Victor Oct 5 '12 at 21:41
The log function is of no great usefulness if you are looking for a closed form solution. – André Nicolas Oct 5 '12 at 22:29
@AndréNicolas - what is your solution? – Victor Oct 5 '12 at 23:06
up vote 2 down vote accepted

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.

share|cite|improve this answer

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 $$ x = -\frac{W(-\frac{a \ln p}{c}p^{b-\frac{ad}{c}})}{a\ln p} - \frac{d}{c} $$

share|cite|improve this answer
... and every branch of $W$ can be used, but at most two will be real. A plot of $x e^x$ will show you that $x e^x = y$ has no real solutions if $y < -1/e$, one (the "principal branch" of $W(y)$) if $y = -1/e$ or $y \ge 0$, and two (the principal branch and the "$-1$" branch) if $-1/e < y < 0$. – Robert Israel Oct 5 '12 at 23:28
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). – Robert Israel Oct 5 '12 at 23:36

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 $x$)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.