Answer:
Here, the given equations,
and
![x^3=12](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ri209hhmbf52u3z03e4uaoawsi2s86mr3w.png)
Similarity: In both equations there is only one variable ( i.e. x)
Difference:
is an exponential equation while
is a polynomial equation.
Now, when we solve an exponential equation we take log in both sides of the equation as follows:
![3^x=12](https://img.qammunity.org/2021/formulas/mathematics/middle-school/qj997438o5795hp6t3kz9zgbh1vp410n1a.png)
![\log(3^x)=\log12](https://img.qammunity.org/2021/formulas/mathematics/middle-school/js53w11otco46rl0miju87q8eh239vt3jy.png)
( ∵
)
![\implies x =(\log 12)/(\log 3)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/qsl1tfxxx8m2cx4agaczxauaauemkchk1r.png)
Hence, the solution of the equation
is
.
While, when we solve a polynomial we find the roots as follows:
![x^3=12](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ri209hhmbf52u3z03e4uaoawsi2s86mr3w.png)
![x^3-12=0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/gt6donas6eh69gq8t6bxdmnwkgjk2g6k86.png)
![x^3-(12^(1)/(3))^3=0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/1rzhztzxeeh9tlx6bnguam5cs0vtz3o00x.png)
![(x-12^(1)/(3))(x^2+12^(1)/(3)x+(12^(1)/(3))^2)=0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ep53tcioi93f9x0drtqfwzt6g1pl3087l4.png)
By zero product property,
or
![(x^2+12^(1)/(3)x+(12^(1)/(3))^2)=0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/xufw3sdcktoqkrd50otbaqt272e2vnsnb1.png)
If
, then
![x=12^(1)/(3)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/uzqh8w15dk5h0gvoszehb6virtxrto5jqy.png)
If
,
Then, by quadratic formula,
![x=\frac{-12^(1)/(3)\pm \sqrt{12^(2)/(3)-4(1)(12^(1)/(3))^2}}{2}](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ujwy2jqbodk7hfmofg42gh9n9v7z86nfot.png)
![=\frac{-12^(1)/(3)\pm \sqrt{12^(2)/(3)-4(12^(2)/(3))}}{2}](https://img.qammunity.org/2021/formulas/mathematics/middle-school/sqcv7257hi7x0tnecef541351taqaie0bu.png)
![=\frac{-12^(1)/(3)\pm i\sqrt{3(12^(2)/(3))}}{2}](https://img.qammunity.org/2021/formulas/mathematics/middle-school/qc7lmuqx7xwqq6sbdtif2j6rsvo7xxkb4h.png)
![=12^(1)/(3)((-1\pm i√(3))/(2))](https://img.qammunity.org/2021/formulas/mathematics/middle-school/bjrnk166y1a28r6b3c8g8758yn1ydntrpm.png)
Hence, the solutions of the equation
are
,
and
.