Solve Equation: Find Real $a$ for $10^a+12^a-14^a=13^a-11^a$

[anemone]

Find all real numbers $a$ for which $10^a+12^a-14^a=13^a-11^a$.

 

[kaliprasad]

Find all real numbers $a$ for which $10^a+12^a-14^a=13^a-11^a$.

Spoiler

a= 2

is a aolution

Spoiler

Take $f(a) = 14^a + 13^a – 12^a – 11 ^a – 10 ^a
= (14^a – 12^a) + (13^a – 11^a ) – 10^a$
We have $14^3 – 12^3 > 10^3$ and gap increases for a >=3 the expression is positive
So we need to look for value < 3
Check for 0 , 1, 2 and we see that a =2 is the integer solution
It may have some non integer solution

 

[chisigma]

Find all real numbers $a$ for which $10^a+12^a-14^a=13^a-11^a$.

[sp]Let's consider the function...

$\displaystyle f(x) = 10^{x} + 11^{x} + 12^{x} - 13^{x} - 14^{x}\ (1)$

By inspection we find easily that (1) vanishes for x=2. For x>2 the negative terms of (1) are dominating, so that thye function sharply decreases. For x<2 the positive terms of (1) are dominating so that is $\lim_{x \rightarrow - \infty} f(x) = 0$ and everywhere is f(x) > 0. The conclusion is that x=2 is the only zero of f(x)...[/sp]

Kind regards

$\chi$ $\sigma$

 

[juantheron]

Find all real numbers $a$ for which $10^a+12^a-14^a=13^a-11^a$.

Solution:

Spoiler

Given $10^a+11^a+12^a = 13^a+14^a$

Now Divide both side by $(12.5)^a$, where $(11.5)^a>0\forall a\in \mathbb{R}$

So $\displaystyle \left(\frac{10}{12.5}\right)^a+\left(\frac{11}{12.5}\right)^a+\left(\frac{12}{12.5}\right)^a = \left(\frac{13}{12.5}\right)^a+\left(\frac{14}{12.5}\right)^a$

So Here $\bf{L.H.S}$ is a sum of strictly Decreasing function while $\bf{R.H.S}$ is a sum of strictly increasing function.

So these two exponential curves intersect each other exactly at one point

So by inspection we get $a = 2$ only solution.

 

[CellCoree]

To solve this equation, we can use the fact that $a^0=1$ for any real number $a$. Therefore, we can rewrite the equation as $(10^a-11^a)+(12^a-13^a)-(14^a-14^0)=0$. Using the properties of exponents, we can simplify this to $10^a(1-1)+12^a(1-1)-14^a(1-1)=0$. This simplifies further to $0=0$, showing that the equation is true for all real numbers $a$. Therefore, there is an infinite number of solutions for this equation.