Need help using scientific calculator

[physicsgal]

-81^(1/4)
= -4square root sign 81
=-3

how do i key this into my calculator:
= "-4square root sign 81"

*note I am working with a $20 casio*

thanks

~Amy

 

[neutrino]

(-)

81

^

0.25

 

[physicsgal]

i see how to work it that way.. but i can't type in "-4square root sign 81", or is that not possible? (im just trying to follow how to do they do it in the book.. i realize both things lead to the right answer)

also

(49x^8y^-2z^6)^(1/2)
=7x^4y-1z^3
=7x^6 * (1/y) * z^3

where does the ^6 come from, or is that a typo in my lesson book?

~Amy

 

[verty]

It depends if your calculator has DAL or not. With an old style calculator, you'd need to enter "8, 1, +/-, y^x, (, 1, /, 4, ), =", with a DAL one you'd enter "-, 8, 1, y^x, (, 1, /, 4, ), =".

 

[physicsgal]

ok

here's another one 9,765,625 = 5^n
n = ?

*n = 10*.. but how would i have figured this out using my calculator?

~Amy

 

[Sane]

That requires the use of the "log" button.

n = log 9765625 / log 5
n = 10[/color]​

The generalization of this rule is...

If ... x^n = y,
Then ... n = log y / log x.​

This is taught in Grade 12 where I live, so I can assume that you don't need to worry about why or how.

 

[physicsgal]

thank you! it works. funny the lesson book doesn't mention log calculations.

~Amy

 

[HallsofIvy]

Please, please, please! Do NOT use "4 square root" for fourth root. It confuses you as well as the people you are talking to (and has been known to cause some math teachers to foam at the mouth).

 

[physicsgal]

Please, please, please! Do NOT use "4 square root" for fourth root. It confuses you as well as the people you are talking to (and has been known to cause some math teachers to foam at the mouth).

thanks, i ll try to remember.

here's one involving exponents.. is my answer in the lowest term or is there more that could be done?

((2^-3 + 5^0)/ (2^-5))^(1/2)

= ((9/8)^(1/2))/ (1/32)

~Amy

 

[Sane]

Not at all, you can reduce it completely. Also, your second step is incorrect as you forgot to square root the denominator. Here's my solution:

$$\begin{equation*}<br /> \begin{split}<br /> (\frac{2^{-3} + 5^{0}}{2^{-5}})^{\frac{1}{2}} &= \frac{\frac{9}{8}^{\frac{1}{2}}}{\frac{1}{32}^{\frac{1}{2}}}\\<br /> \\<br /> &= (\frac{3}{\sqrt{8}}) (\frac{\sqrt{32}}{1})\\<br /> \\<br /> &= \frac{3\sqrt{4}\sqrt{8}}{\sqrt{8}}\\<br /> \\<br /> &= 6<br /> \end{split}<br /> \end{equation*}$$

 

[physicsgal]

thanks. i did it 'the amy way' and got the same answer. just converted the numbers to normal numbers (1.125/0.03125)^(1/2).

also, is there any way to convert this to something like this: (so you can just figure it out be crossing out the like numbers):

[((1/2)(1/2)(1/2) + 1)/ ((1/2)(1/2)(1/2)(1/2)(1/2))] ^ (1/2)

= [1/(1/2)(1/2)]^(1/2)??

with that i end up with = 2. why is that method not working??

~Amy

 

[Sane]

No... $\frac{a + b}{(a)(c)}$ is not the same as $\frac{b}{c}$. Although that's what your solution suggests; you cancel out the (1/2), even though you have two separate terms in the numerator. It doesn't make sense to be able to cancel out the a's when their weight on the equation is inequivilent.

I suggest that you practice simplifying these algebraic equations (as I showed in my last post), learn the rules of Math, and prevent yourself from following these techniques blindly. If you don't fix your presumptious tuitions of what you can do to an equation, you're going to run into some painful potholes down the rode.

 

[physicsgal]

thanks for the tips. i somewhat see what you're saying. i should go over my lesson books again. I am doing this as a study-at-home course so that's part of the problem (me teaching myself)

i appreciate the help.

~Amy