# Which point satisfies both f(x)=2^x and g(x)=3^x?

Nov 17, 2016

$\left(0 , 1\right)$

#### Explanation:

If $f \left(x\right) = y = g \left(x\right)$ then we have:

${2}^{x} = {3}^{x}$

Divide both sides by ${2}^{x}$ to get:

$1 = {3}^{x} / {2}^{x} = {\left(\frac{3}{2}\right)}^{x}$

Any non-zero number raised to the power $0$ is equal to $1$. Hence $x = 0$ is a solution, resulting in:

$f \left(0\right) = g \left(0\right) = 1$

So the point $\left(0 , 1\right)$ satisfies $y = f \left(x\right)$ and $y = g \left(x\right)$

Note also that since $\frac{3}{2} > 1$, the function ${\left(\frac{3}{2}\right)}^{x}$ is strictly monotonically increasing, so $x = 0$ is the only value for which ${\left(\frac{3}{2}\right)}^{x} = 1$