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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?

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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.

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

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... 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 $x$)

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