# Exponential equation

1. Nov 5, 2013

### juantheron

No. of Real solution of the equation $2^x\cdot \ln (2)+3^x\cdot \ln (3)+4^x\cdot \ln (4) = 2x$

. The attempt at a solution

Let $f(x) = 2^x\cdot \ln (2)+3^x\cdot \ln (3)+4^x\cdot \ln (4) -2x$

Here solution must be exists for $x>0$, because for $x\leq 0$, L.H.S>0 while R.H.S <0

$f^{'}(x) = 2^x\cdot \left(\ln 2\right)^2+3^x\cdot \left(\ln 3\right)^2+4^x\cdot \left(\ln 4\right)^2 - 2$

and $f^{''}(x) = 2^x\cdot \left(\ln 2\right)^3+3^x\cdot \left(\ln 3\right)^3+4^x\cdot \left(\ln 4\right)^3 >0\forall x\in \mathbb{R}$

Means function $f(x)$ is Concave - Upward and $f^{'}(x)$ is Strictly Increasing function

and when $x\rightarrow +\infty, f^{'}(x)\rightarrow \infty$ and when $x\rightarrow -\infty, f^{'}(x)\rightarrow -\infty$

So using LMVT, function $f^{'}(x) = 0$ has exactly one real root.

But I did not understand How can i calculate root of $f(x) = 0$

2. Nov 5, 2013

### Mentallic

There isn't a real root, but even if there were, this analysis won't give you the value of the root(s).

Can you show that f'(x) crosses the x-axis only once? And also, since you've already shown that f''(x) > 0 for all x hence f(x) is concave upwards, what can you say about f(x) where x is the turning point (where f'(x)=0)?

If you can't quite understand this, picture a general parabola that's concave upwards. It's essentially the same problem.

3. Nov 5, 2013

### Ray Vickson

The function g(x) = ln(2)*2^x + ln(3)*3^x + ln(4)*4^x - 2*x is strictly convex on the real line (what you call concave up --- an archaic terminology no longer used by optimization people), so there are either 0, 1 or 2 real solutions of g(x) = 0. What you checked is not sufficient to establish the existence of a real root; can you really say for sure that there must be a root?