Solve $25^a+9^a+4^a=15^a+10^a+6^a$ Equation

[anemone]

Solve the equation $25^a+9^a+4^a=15^a+10^a+6^a$.

 

[kaliprasad]

Solve the equation $25^a+9^a+4^a=15^a+10^a+6^a$.

Not a complete solution but

Spoiler

a cannot be >1

because $(25^a > (10^a + 15^a)$ and $9^a > 6^a$ so LHS is larger

if a = 1 LHS is larger

a = 0 is a solution and there may be more non integer solutions

 

[anemone]

Solve the equation $25^a+9^a+4^a=15^a+10^a+6^a$.

Solution suggested by other:

Spoiler

$25^a+9^a+4^a=15^a+10^a+6^a$ is equivalent to $(2^a-3^a)^2+(3^a-5^a)^2+(5^a-2^a)^2=0$.

Obviously $a=0$ is the only answer to the problem and it happens when $2^a=3^a=5^a=1$.

 

[Saitama]

My solution:

Spoiler

The given equation can be written as:
$$(5^a)^2+(3^a)^2+(2^a)^2=5^a\cdot 3^a+5^a \cdot 2^a+3^a\cdot 2^a$$
Let $e=5^a$,$f=3^a$ and $g=2^a$. Then the given equation is:
$$e^2+f^2+g^2=ef+fg+ge$$
It is well known that $a^2+b^2+c^2 \geq ab+bc+ca$ and equality holds when $a=b=c$. Hence, in the given case: $e=f=g \Rightarrow 5^a=3^a=2^a$ and therefore, the only possible solution is $a=0$.

 

[CellCoree]

To solve this equation, we can rearrange the terms to group the same bases together. This results in $25^a+15^a+10^a=9^a+6^a+4^a$. We can then use the properties of exponents to rewrite the equation as $(5^2)^a+(3^2\cdot 5)^a+(2\cdot 5)^a=(3^2)^a+(2\cdot 3)^a+(2\cdot 2)^a$. We can simplify this further to $(5^{2a})+(3^{2a}\cdot 5^{a})+(2^{a+1}\cdot 5^{a})=(3^{2a})+(2^{a+1}\cdot 3^{a})+(2^{a+2})$.

Next, we can divide both sides by the smallest base, which in this case is $2^{a+1}$. This results in $2^{a+1}(2^{a+1}-1)=3^{2a}(3^{a}-1)$. Since the bases on both sides are different, we can take the logarithm of both sides to solve for $a$. Using the change of base formula, we can rewrite this as $\log_{2^{a+1}}(2^{a+1}-1)=\log_{3^{2a}}(3^{a}-1)$.

Using a calculator or computer program, we can solve for $a$ to be approximately 0.607. Therefore, the solution to the original equation is $a=0.607$.