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

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The equation 10^a + 12^a - 14^a = 13^a - 11^a is analyzed through the function f(x) = 10^x + 11^x + 12^x - 13^x - 14^x. It is determined that f(x) equals zero at x = 2. For values greater than 2, the function decreases sharply due to the dominance of negative terms, while for values less than 2, positive terms dominate, keeping f(x) positive. Thus, x = 2 is identified as the only solution to the equation. The analysis concludes that the only real number a satisfying the equation is a = 2.
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Find all real numbers $a$ for which $10^a+12^a-14^a=13^a-11^a$.
 
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anemone said:
Find all real numbers $a$ for which $10^a+12^a-14^a=13^a-11^a$.

a= 2
is a aolution

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
 
anemone said:
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

$\chi$ $\sigma$
 
anemone said:
Find all real numbers $a$ for which $10^a+12^a-14^a=13^a-11^a$.

Solution:

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.
 
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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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