3. The solutions to the equation $\frac{\log \left(35-x^{3}\right)}{\log (5-x)}=3$ are the lengths of the legs of a right triangle. Calculate the radius of the circle circumscribed around this triangle.
3. We multiply the equation by $\mathrm{z} \log (5-x)$, antilogarithmize and rearrange into a quadratic $15 x^{2}-75 x+90=0$. The solutions to the quadratic equation are $x_{1}=2, x_{2}=3$. The solutions to the equation are the lengths of the legs of a right triangle. We calculate the length of the hypotenuse. Half the length of the hypotenuse is equal to the radius $\mathrm{R}$ of the circle circumscribed around the triangle.

Rearranging the equation after antilogarithmizing $\log \left(35-x^{3}\right)=3 \log (5-x) \ldots \ldots \ldots \ldots \ldots .1$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_2b0bf3255e176946d93cg-09.jpg?height=55&width=1639&top_left_y=2114&top_left_x=206)

![](https://cdn.mathpix.com/cropped/2024_06_07_2b0bf3255e176946d93cg-09.jpg?height=48&width=1639&top_left_y=2169&top_left_x=206)

![](https://cdn.mathpix.com/cropped/2024_06_07_2b0bf3255e176946d93cg-09.jpg?height=57&width=1639&top_left_y=2210&top_left_x=206)

Calculating the radius $R=\frac{\sqrt{13}}{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .1$ point