Solve the equation $2^x+3^x-4^x+6^x-9^x=1$

Question

I found an interesting equation I don't seem to grasp completely.

Solve the equation:

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

Now. I thought the only relevant solution was the integer solutions, but I was sorely mistaken.

So if the question was: Find all integer solutions to the following equation my attempt is as follows:

$2^x+3^x-4^x+6^x-9^x=1$ (1)

$2^x+3^x+3^x2^x-(2^x)^2-(3^x)^2=1$

Substitution:

$P=2^x\quad$ $Q=3^x$ (2)

$P+Q+PQ-P^2-Q^2=1$

We can see that this equation is symmetric and that $P>0, Q>0$

For $P=1$ we get $1+Q+Q-1-Q^2=1 \to Q=1$

P & Q are equal, with (2) we get: $P=Q \to 2^x=3^x \to x=0$

Answer: $x=0$ is the only solution

Now when I wrote down my solution I recognize that several steps are wrong. First of all how do i know P and Q are equal? They might not be i suppose

If anyone could give me a hint on how to solve these type of equations or even how to solve this equation I'd be very grateful.

And furthermore is this type of equation called exponential equation?

The type of exponential equation I'm more familiar with is simply solved with log rules, equations were you can simplify both sides and just log and solve.

Answer 1

Note that the equation is equivalent to $$(2^x-1)^2-(2^x-1)(3^x-1)+(3^x-1)^2=0.$$ Moreover, note that the above equation is satisfied iff $x=0$ because if $v\not=0$ then $$u^2-uv+v^2=v^2((u/v)^2-(u/v)+1)>0.$$

Answer 2

Given conic section $$ P^2 - PQ + P^2 - P - Q + 1 = 0, $$ the $P$ partial derivative is $$ 2P-Q - 1. $$ The $Q$ partial is $$ -P + 2Q - 1. $$ Setting both to zero gives $P=Q=1.$

So $$ P = u + 1, \; \; Q = v + 1. $$

$$ P^2 - PQ + P^2 - P - Q + 1 = u^2 - uv + v^2. $$ The set $u^2 - uv + v^2 = 0$ is a single point $u=v=0,$ as this is a positive quadratic form: the Hessian matrix is positive.

Oh, if the Hessian has determinant zero, there will not be a "center," the thing is a parabola or degenerate parabola.