http://img11.imageshack.us/img11/8986/scan0002lq.jpg

i understood a,b and c but d made not alot of sense. how do i go backwards? thanks

Printable View

- Mar 8th 2010, 04:31 PMminneola24Backward Exponets
http://img11.imageshack.us/img11/8986/scan0002lq.jpg

i understood a,b and c but d made not alot of sense. how do i go backwards? thanks - Mar 8th 2010, 04:36 PMQuacky
Hint:

$\displaystyle \sqrt x = x^\frac{1}{2}$ - Mar 8th 2010, 04:38 PMminneola24
still confused.

- Mar 8th 2010, 04:39 PMQuacky
$\displaystyle \sqrt{64}=8$.

So, using what I've said above, to what power must you raise 64 to get 8? - Mar 8th 2010, 04:48 PMminneola24
64 ^ .5 = 8

thanks

i cant do 64 ^ .5 to get 2 so what do I do? - Mar 8th 2010, 04:53 PMQuacky
It's better to think about the exponents as fractions.

If $\displaystyle 2^2=4$

Then $\displaystyle 4^\frac{1}{2}=2$

If $\displaystyle 9^4=6561$

Then $\displaystyle 6561^\frac{1}{4}=9$

So it's easiest to work backwards.

$\displaystyle 3^4=81$

So $\displaystyle 81^?=3$ - Mar 8th 2010, 04:54 PMVertazontal
For G, H and I remember that an exponent raised to an exponent is just the two exponents multiplied together.

so,

x^2^6=x^12 - Mar 8th 2010, 05:07 PMminneola24
so to get 81 to 3 you do 81^.25

hjowever for 64 -> 2 how do i do 2^1/6 to check? - Mar 8th 2010, 05:12 PMQuacky
That's correct! (Happy)

- Mar 8th 2010, 05:15 PMminneola24
awsome, thank you.

http://img197.imageshack.us/img197/2392/scan0002iu.jpg

i know its unrelated, but how do i do these?

for a i have

.5 = x

is that correct? thanks - Mar 8th 2010, 05:24 PMQuacky
$\displaystyle 125^x=25^{x+1}$

I would do that by writing both as 5^{something}

$\displaystyle 125=5^3$

$\displaystyle 25=5^2$

So $\displaystyle (5^3)^x=(5^2)^{x+1}$

$\displaystyle 5^{3x}=5^{2x+2}$

So:

$\displaystyle 3x=2x+2$

$\displaystyle x=2$

Part b) is harder. Try to write it out as powers of 3.

For part c) Try to write it out as powers of 2 - Mar 8th 2010, 05:25 PMintegral
Is this not a logarithm problem?

$\displaystyle 3^x=27 \therefore log_3 27=x$

then

$\displaystyle x=\frac{log27}{log3}$ - Mar 8th 2010, 05:27 PMVertazontal
- Mar 8th 2010, 07:21 PMBacterius
Don't confuse the OP, integral :o (even your signature seems to be an extension of your answer xD)

For part c), you could note that $\displaystyle 4 = 2 \times 2$, and try to find yourself with just two's on the right side. Then note that $\displaystyle 8 = 2^3$. The rest should follow.