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?
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?
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.
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} $
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 $x$)