The curve with equation $y=3x^2$ meets the curve with equation $y=15-2^{x}$ at the point $P$ - Edexcel - A-Level Maths Pure - Question 7 - 2020 - Paper 2

Now we can rewrite the equation as:

$2^{x} = 15 - 3x^2$

Next, we can test integer values for $x$ to find a solution. Let's try $x = 3$:

$2^{3} = 8 ext{ and } 15 - 3(3^2) = 15 - 27 = -12 ext{ (not a solution)}$

Now, let's try $x = 2$:

$2^{2} = 4 ext{ and } 15 - 3(2^2) = 15 - 12 = 3 ext{ (not a solution)}$

Next, we evaluate $x = 1$:

$2^{1} = 2 ext{ and } 15 - 3(1^2) = 15 - 3 = 12 ext{ (not a solution)}$

Lastly, for $x = 0$:

$2^{0} = 1 ext{ and } 15 - 3(0^2) = 15 ext{ (not a solution)}$

Using logarithms, we can rearrange to express $x$ in terms of known values:

From $3x^2 + 2^{x} = 15$, we can also solve using properties of logarithms if we manipulate equations further, leading to possible values of $x = ext{log}_2(15 - 3x^2)$, but this should ideally yield $x= ext{log}_2(3)$ through further manipulation.