1700 Followers Problem

Let $\space LHS = y = 1700^{\log_A{17}}$

$\Rightarrow \log_A{y} = \log_A{\left( 1700^{\log_A{17}} \right)} = \log_A{17} (\log_A{1700} ) = \log_A{\left(17^{\log_A{1700}}\right)} \\ \Rightarrow y = 17^{\log_A{1700}} = RHS$

$\Rightarrow 1700^{\log_A{17}} - 17^{\log_A{1700}} = \boxed{0}$

Simply solve first the roots of the first equation. It can be done simply since it is reducible in quadratic form. Then it can be found that one of the roots is 5 , so substituting it to the second equation, 5 also satisfies it hence the answer must be 0

No matter the base, we always have: $a^{\log b}=b^{\log a}$ Proof: Writing $e$ for the base (though it really could be anything), it's easy to see that they are both equal to $e^{(\log a)(\log b)}$ . $\quad\square$

This immediately gives us $1700^{\log_A17}=17^{\log_A1700}$ , which means that $\boxed{1700^{\log_A17}-17^{\log_A1700}=0}$ .

$\large \color{#D61F06}{ 1700^{\log_A{17} } } -\color{#3D99F6}{17^{\log_A{1700}} } \\Taking ~\log_A:-~~of ~each~ term.\\ \color{#D61F06}{\log_A{17} *\log_A{1700}..Left~ term.} ......\color{#3D99F6}{\log_A{1700}*\log_A{17}..Right~term.}\\\text{Log to the same base are equal, so the terms are also equal.}\\\text{So, expression=0.}\\\text{No need of the given two equations. }$

As mentioned by another solution, it can be shown that $x^{\log_y z} = z^{\log_y x}$ , so you don't even need any other information to know the result is $0$ .

That said, if you did want to find the roots as an exercise, then I'd recommend starting with $x^3 - 6x^2 + 5x = x(x-5)(x-1) = 0$ . Since the constant term of the second equation is $45$ , it's worth verifying that $x-5$ is a root. Once you discover that it is, you can factor out the rest: $2x^4 - 11x^3 + 23x^2 - 99x + 45 = (x-5)(2x-1)(x^2+9) = 0$ .