Since the exponent is $x/2$, a change of $\Delta x$ changes the exponent by $\Delta x/2$. Find the exponent changes for $\Delta x=6$ and for $\Delta x=5$.

You can write $b^{5/2}$ in terms of $b^3$ by using $b^{5/2}=(b^3)^{(5/2)/3}$. Then substitute $b^3=1+\frac{p}{100}$ and convert back to a percent with $(k-1)\cdot 100$.

In Desmos, enter f(x)=240*b^(x/2) and create sliders for b (choose values > 1) and x. ©⁠ Aniko

Enter p=100*(f(x+6)-f(x))/f(x).

Enter c_actual=100*(f(x+5)-f(x))/f(x).

Enter c_formula=((1+p/100)^(5/6)-1)*100 and verify that c_formula matches c_actual for different values of b and x.

When $x$ increases by 6, the exponent $x/2$ increases by $6/2=3$.

So

If $f(x)$ increases by $p$%, then

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When $x$ increases by 5, the exponent $x/2$ increases by $5/2$, so

Use exponent rules to connect $b^{5/2}$ to $b^3$:

Substitute $b^3=1+\frac{p}{100}$:

If a quantity is multiplied by $k$, then the percent increase is $(k-1)\cdot 100$.

Here $k=\left(1+\frac{p}{100}\right)^{5/6}$, so