## burhan101 2 years ago Solve for 'x'

1. burhan101

$\huge 0=1- \frac{ 1 }{ 2x^{\frac{ 3 }{ 2}}}$

2. Abhishek619

|dw:1371705222820:dw|

3. burhan101

how does it equal one ?!

4. terenzreignz

slowly :D Why don't we bring $$\huge \frac1{2x^{\frac32}}$$ to the left side like this... $\Large \frac1{2x^{\frac32}}=1$ clear now? :)

5. burhan101

|dw:1371743475441:dw|

6. burhan101

yes

7. terenzreignz

We basically added $\huge \frac1{2x^{\frac32}}$ to both sides as per the addition property of equality.

8. terenzreignz

Well then, can you solve it from here?

9. burhan101

do i multiply by both sides?

10. terenzreignz

Sure :)

11. burhan101

|dw:1371743682268:dw|

12. terenzreignz

Never mind that for now. Why don't we multiply both sides by $\Large 2x^{\frac32}$ That gives us... $\Large 1 = 2x^{\frac32}$ And divide both sides by 2...

13. burhan101

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

14. terenzreignz

true dat :) Now you can raise both sides to the $$\LARGE \frac23$$ power (just to get rid of the exponent of x) You then get...?

15. burhan101

|dw:1371743929750:dw| like this?

16. terenzreignz

That gives you x. What about the right-hand side?

17. burhan101

0/63

18. burhan101

0.63 *

19. terenzreignz

come again? :D

20. terenzreignz

Oh, were you supposed to give it in decimal form? okay then :D

21. burhan101

I can get a fraction answer too?

22. terenzreignz

Well, the best you can hope for is the ugly-looking $\Large \left(\frac12\right)^{\frac23}$

23. terenzreignz

and that is the correct answer :)

24. burhan101

oh so i should probably keep it,

25. burhan101

thank youuu !!!

26. terenzreignz

Yeah, unless you were specifically asked for an approximation :)

27. terenzreignz

Oh well... now I have to go :) Please practice solving algebraic equations :D ----------------------------------------- Terence out