$2^x + 3^x - 4^x + 6^x - 9^x ≤ 1$ $\forall x \in R$

Question

How can i prove that $2^x + 3^x - 4^x + 6^x - 9^x ≤ 1$ $\forall x \in R$.

I tried $log(2^x + 3^x - 4^x + 6^x - 9^x) = log (1296^x) = x log(1296)$ i don't know if im correct here i stuck some help please

$\ln\left(2^x + 3^x - 4^x + 6^x - 9^x\right)\neq\ln\left(1296^x\right).$ — Hakim, Jun 18 '14 at 20:06
The logarithm of a sum or difference does not have any simple general expression. — colormegone, Jun 18 '14 at 20:26

score 14 · Accepted Answer · answered Jun 18 '14 at 20:06

14

Note that:

$9^x-6^x+4^x-3^x-2^x+1$

$= 9^x-2 \cdot 6^x + 4^x + 6^x-3^x-2^x+1$

$= \left[(3^x)^2-2(3^x)(2^x) + (2^x)^2\right] + \left[(3^x)(2^x)-3^x-2^x+1\right]$

$= (3^x-2^x)^2 + (3^x-1)(2^x-1)$

$\ge 0$

since $3^x-1$ and $2^x-1$ have the same sign.

Therefore, $2^x+3^x-4^x+6^x-9^x \le 1$

answered Jun 18 '14 at 20:06

JimmyK4542

How do you prove they have the same sign? – Elliot Gorokhovsky Jun 19 '14 at 20:37
If $x > 0$, then $2^x > 1$ and $3^x > 1$. If $x < 0$, then $2^x < 1$ and $3^x < 1$. If $x = 0$, then $2^x = 1$ and $3^x = 1$. – JimmyK4542 Jun 19 '14 at 20:41

score 4 · Answer 2 · answered Jun 18 '14 at 20:36

4

Note that $$9^x-6^x+4^x-3^x-2^x+1=\frac{1}{2}\left(( 3^x-1)^2+(2^x-1)^2+(3^x-2^x)^2\right)>0$$

answered Jun 18 '14 at 20:36

Omran Kouba

2 Answers2