Respuesta :
Answer:
Here, the given equations,
[tex]3^x=12[/tex] and [tex]x^3=12[/tex]
Similarity: In both equations there is only one variable ( i.e. x)
Difference: [tex]3^x=12[/tex] is an exponential equation while [tex]x^3=12[/tex] is a polynomial equation.
Now, when we solve an exponential equation we take log in both sides of the equation as follows:
[tex]3^x=12[/tex]
[tex]\log(3^x)=\log12[/tex]
[tex]x\log 3 =\log 12[/tex] ( ∵ [tex]\log m^n=n\log m[/tex] )
[tex]\implies x =\frac{\log 12}{\log 3}[/tex]
Hence, the solution of the equation [tex]3^x=12[/tex] is [tex]x=\frac{\log 12}{\log 3}[/tex].
While, when we solve a polynomial we find the roots as follows:
[tex]x^3=12[/tex]
[tex]x^3-12=0[/tex]
[tex]x^3-(12^\frac{1}{3})^3=0[/tex]
[tex](x-12^\frac{1}{3})(x^2+12^\frac{1}{3}x+(12^\frac{1}{3})^2)=0[/tex]
By zero product property,
[tex](x-12^\frac{1}{3})=0[/tex] or [tex](x^2+12^\frac{1}{3}x+(12^\frac{1}{3})^2)=0[/tex]
If [tex](x-12^\frac{1}{3})=0[/tex], then [tex]x=12^\frac{1}{3}[/tex]
If [tex]x^2+12^\frac{1}{3}x+(12^\frac{1}{3})^2=0[/tex],
Then, by quadratic formula,
[tex]x=\frac{-12^\frac{1}{3}\pm \sqrt{12^\frac{2}{3}-4(1)(12^\frac{1}{3})^2}}{2}[/tex]
[tex]=\frac{-12^\frac{1}{3}\pm \sqrt{12^\frac{2}{3}-4(12^\frac{2}{3})}}{2}[/tex]
[tex]=\frac{-12^\frac{1}{3}\pm i\sqrt{3(12^\frac{2}{3})}}{2}[/tex]
[tex]=12^\frac{1}{3}(\frac{-1\pm i\sqrt{3}}{2})[/tex]
Hence, the solutions of the equation [tex]x^3=12[/tex] are [tex]12^\frac{1}{3}[/tex], [tex]12^\frac{1}{3}(\frac{-1+i\sqrt{3}}{2})[/tex] and [tex]12^\frac{1}{3}(\frac{-1-i\sqrt{3}}{2})[/tex] .
