Finding x using logarithms

Question

$$x^{1/3} - 4x^{-1/3}= 3$$

I dont quite understand what to do with the 4 in the term 4x so that I can rewrite the two terms with the same base.

Also since there are two terms how do i solve this? I'm supposed to find x

I don't think you are supposed to solve this using logarithms, and it is also very unclear. Please use MathJax to clear it up. — Simply Beautiful Art, May 21 '17 at 17:18

score 2 · Answer 1 · answered May 21 '17 at 17:26

2

If you multiply by $x^{1/3}$ you get

$$x^{2/3}-4 = 3x^{1/3}$$

Then

$$x^{2/3}-3x^{1/3} -4 = (x^{1/3} -4)(x^{1/3}+1) = 0$$

which gives you two solutions. Logarithms aren't needed.

answered May 21 '17 at 17:26

B. Goddard

1

score 0 · Answer 2 · answered May 21 '17 at 17:22

0

Put $x^(1/3)=t$;

=>$t-4*t^(-1)=3$;

=>$t-4/t=3;$

=>$t^2-3t-4=0;$

=>$(t+1)(t-3)=0;$

=>$t=3 or -1$

=>$x=-1 or 9$

answered May 21 '17 at 17:22

Vikranth Inti

119

$(t+1)(t-4)=0$ might work better. So too might cubing at the end – Henry May 22 '17 at 07:03

score 0 · Answer 3 · answered May 21 '17 at 17:26

You can solve it in the following way: 1) Rewrite the equation by substituting $y=x^\frac{1}{3}$. Note that then $\frac{1}{y}=x^{-\frac{1}{3}}$. 2) Multiply the equation by $y$. 3) Now you have a quadratic equation in $y$. Solve it using the usual formula. 4) When you have found the solutions, substitute back (ie. write $x^\frac{1}{3}$ instead of $y$). Then the complete solution is straightforward.

Finding x using logarithms

3 Answers3