Exponential equation

  • Thread starter juantheron
  • Start date
  • #1
21
1
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##

Help please

Thanks in Advance.
 

Answers and Replies

  • #2
Mentallic
Homework Helper
3,798
94
and when ##x\rightarrow -\infty, f^{'}(x)\rightarrow -\infty ##
Are you sure about that?

But I did not understand How can i calculate root of ##f(x) = 0##
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
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
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##

Help please

Thanks in Advance.


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?
 

Related Threads on Exponential equation

  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
6
Views
7K
  • Last Post
Replies
1
Views
993
Replies
1
Views
1K
Replies
3
Views
5K
Replies
5
Views
3K
Replies
5
Views
4K
Top