# How do you solve 3 log x = 6 - 2x?

Jan 13, 2016

Not sure if it can be solved

$x = 2.42337$

#### Explanation:

Other than using Newton's method, I am not sure if it is possible to solve this. One thing you can do is prove that it has at exactly one solution.

$3 \log x = 6 - 2 x$
$3 \log x + 2 x - 6 = 0$

Set:

$f \left(x\right) = 3 \log x + 2 x - 6$

Defined for $x > 1$

$f ' \left(x\right) = \frac{3}{x \ln 10} + 2$

$f ' \left(x\right) = \frac{3 + 2 x \ln 10}{x \ln 10}$

For every $x > 1$ both the numerator and denominator are positive, so the function is increasing. This means it can only have a maximum of one solution (1)

Now to find all the values of $f \left(x\right)$ $x > 1$ means $x \in \left(0 , \infty\right)$:

${\lim}_{x \to {0}^{+}} f \left(x\right) = {\lim}_{x} \to \left({0}^{+}\right) \left(3 \log x + 2 x - 6\right) = - \infty$

${\lim}_{x \to \infty} f \left(x\right) = {\lim}_{x \to \infty} \left(3 \log x + 2 x - 6\right) = \infty$

Therefore, $f \left(x\right)$ can take any real value, including 0, which means that $f \left(x\right) = 0 \iff 3 \log x + 2 x - 6 = 0$ can be a solution at least once (2)

(1) + (2) = (Maximum of one) + (At least one) = Exactly one