Try to express both $2^{x/2}$ and $2^{x/3}$ as powers of the same base, like $2^{x/6}$. Then think about using a substitution such as $a = 2^{x/6}$.

After substituting $a = 2^{x/6}$, you should get an equation involving $a^3$ and $a^2$. Rearrange it so it equals 0, and then look for a value of $a$ that makes the equation true (try small integers).

In the first expression line, enter 2^(x/2) + 2^(x/3) so Desmos graphs the function $y = 2^{x/2} + 2^{x/3}$.

In a new line, enter y = 12 to draw a horizontal line representing the constant value 12.

Use the intersection tool (or tap on the point where the curve and the horizontal line cross). The x-coordinate of this intersection is the solution to the equation $2^{x/2} + 2^{x/3} = 12$.

Start with the equation

Notice that $x/2$ and $x/3$ can both be written with denominator 6:

• $x/2 = 3x/6$
• $x/3 = 2x/6$

So you can rewrite the powers of 2 using $x/6$:

• $2^{x/2} = 2^{3x/6} = \bigl(2^{x/6}\bigr)^3$
• $2^{x/3} = 2^{2x/6} = \bigl(2^{x/6}\bigr)^2$

Let $a = 2^{x/6}$. Then

• $2^{x/2} = a^3$
• $2^{x/3} = a^2$

Substitute into the original equation:

Move all terms to one side:

This is a polynomial (specifically a cubic) equation in $a$.

Test small integer values of $a$ (1, 2, 3, ...). Since $a = 2$ makes $2^3 + 2^2 - 12 = 8 + 4 - 12 = 0$, $(a - 2)$ is a factor. Factor:

Because $a = 2^{x/6} > 0$ and $a^2 + 3a + 6 = 0$ has no real roots, the only valid real solution is $a = 2$. Thus $2^{x/6} = 2$, so $x/6 = 1$ and $x = 6$.