Is this correct?
- Thread starter cosmic
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I get something quite different. Please reply showing your steps, so we can try to help you find the error(s). Thank you!
I get something quite different. Please reply showing your steps, so we can try to help you find the error(s). Thank you!
Thank you for your reply. initially I used the law (am)n=amn to turn (9x3)1/2 to (9x)3/2. Then I used the law (ab)n=anbn. Thus turning (8x)2/3 to 4x2/3 and (9x)3/2 to 27x3/2. I also converted the radical to a rational exponent to get (64x)1/6 and ended up with 2x1/6 after using a similar process as before. Thus I had (10813/6)/(2x1/6) getting 54x2.
Throughout I assumed that, for instance, the term (8x)2/3 consists of two bases 8 and x hence why I used the rule (am)n=amn . I hope this helps you understand the rationale used.
Hi cosmic:
In the given expression above (blue), the factor 9 is not being cubed. In your result (red), you cubed 9. That's a mistake.
These look good. Fix your expression in red above, put the pieces together, and simplify.
Cheers
Hi cosmic:
In the given expression above (blue), the factor 9 is not being cubed. In your result (red), you cubed 9. That's a mistake.
These look good. Fix your expression in red above, put the pieces together, and simplify.
Cheers
Oh I see where I went wrong, thank you so much. I ended up with 6x2 as my final answer. I hope that I haven't made any more stupid mistakes!
D
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(8x)2/3 = (23x)2/3 = 4(x)2/3
(9x3)1/2 = (32x3)1/2 = 3(x)3/2
[√(64x)]1/3 = [(26x)1/2]1/3 = [26x]1/6 = 2[x]1/6
And the answer is 6x2 - you are correct.
No, but you will.
That's because mathematical knowledge proceeds by making mistakes!As long as you have the ability to understand what went wrong, to correct the error, and to move forward, then it's not really a "stupid" mistake. It's a learning experience.
Here is another approach to the same exercise. This approach groups constants and groups variables, and simplifies each group separately.
It also makes use of the property a^m/a^n = a^(m-n)
\(\displaystyle \dfrac{(8x)^{2/3}(9x^3)^{1/2}}{(\sqrt{64x})^{1/3}}\)
\(\displaystyle \dfrac{8^{2/3} 9^{1/2}}{64^{1/6}} \cdot \dfrac{x^{2/3} x^{3/2}}{x^{1/6}}\)
\(\displaystyle \dfrac{4 \cdot 3}{2} \cdot x^{13/6 - 1/6}\)
\(\displaystyle 6x^2\)
No, but you will.
As long as you have the ability to understand what went wrong, to correct the error, and to move forward, then it's not really a "stupid" mistake. It's a learning experience.
Here is another approach to the same exercise. This approach groups constants and groups variables, and simplifies each group separately.
It also makes use of the property a^m/a^n = a^(m-n)
\(\displaystyle \dfrac{(8x)^{2/3}(9x^3)^{1/2}}{(\sqrt{64x})^{1/3}}\)
\(\displaystyle \dfrac{8^{2/3} 9^{1/2}}{64^{1/6}} \cdot \dfrac{x^{2/3} x^{3/2}}{x^{1/6}}\)
\(\displaystyle \dfrac{4 \cdot 3}{2} \cdot x^{13/6 - 1/6}\)
\(\displaystyle 6x^2\)
Thank you so much guys. You've been a massive help. I hope I can learn a few things from you in the coming years.