proof - proving a proposition involving logarithms is true or false

Question

I'm looking at my textbook and I'm not sure how to solve this to prove whether it's true or not.

(there exists x in the real)(3^x = x^2 )

Any help would be good. Thank you.

Answer 1

Hint: To see why such an $x$ exists, apply the Intermediate Value Theorem. Consider the continuous function: $$ f(x) = x^2 - 3^x $$ and the closed interval $[-1,0]$.

Answer 2

A simple curve plot (or even sketch) should convince you of the answer.

But it's also easy to show algebraically. Note that you're looking for zeroes of $f(x) = 3^x - x^2$. Both terms are always non-negative. For all positive values of $x$, $3^x$ is larger than $x^2$ (proving this rigorously can be done by a bit of differential calculus), and $f(x)>0$. For large magnitude negative values of $x$, $3^x \to 0^+$ while $x^2$ returns large positive values, giving $f(x)<0$. Since the function is continuous, the intermediate value theorem guarantees a real zero for $f(x)$.

The actual value cannot be solved for exactly, but it can be estimated by numerical methods or accurate curve plotting. The only real zero is approximately $-0.686$. See: http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIzXngteF4yIiwiY29sb3IiOiIjMDAwMDAwIn0seyJ0eXBlIjoxMDAwfV0-