# How do you solve 12^(x) - 5 > 6 ^(1-x)?

Aug 20, 2015

$x > 0.756214$ this is a numerical approximation to the solution and only accurate to 6 s.f..

#### Explanation:

Move all expressions to one side of the inequality:

$f \left(x\right) = {12}^{x} - 5 - {6}^{1 - x} > 0$

Taking the derivative of the left side with respect to $x$ we have:

$f ' \left(x\right) = \log \left(12\right) {12}^{x} + \log \left(6\right) {6}^{1 - x}$

This expression is positive for all $x \in \mathbb{R}$, therefore $f \left(x\right)$ is strictly increasing at all times. This leaves two possibilities:

1) The expression is always positive and therefore the inequality is satisfied for all $x \in \mathbb{R}$
2) The expression has a single root and satisfies the inequality for any $x$ greater than that root.

If we can find any value of $f \left(x\right)$ that is negative then the second possibility is true. We can see this is true for $f \left(0\right)$:

$f \left(0\right) = {12}^{0} - 5 - {6}^{1 - 0} = - 10$

If we can solve $f \left(x\right) = 0$ then we solve this problem. There is no obvious way I can see to analytically solve this expression so I would use a numerical method. if you use the Newton-Raphson method (https://en.wikipedia.org/wiki/Newton%27s_method) with an initial guess of ${x}_{0} = 0$ then the series:

${x}_{n + 1} = {x}_{n} - f \frac{{x}_{n}}{f ' \left({x}_{n}\right)}$

Will converge on the root. We find that this converges after a few iterations:

$x = 0.756214 -$6 s.f.

Therefore we have the solution:

$x > 0.756214$ knowing this is a numerical approximation to the solution and only accurate to 6 s.f..