# Backward Exponets

• Mar 8th 2010, 04:31 PM
minneola24
Backward 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 PM
Quacky
Hint:
$\sqrt x = x^\frac{1}{2}$
• Mar 8th 2010, 04:38 PM
minneola24
still confused.
• Mar 8th 2010, 04:39 PM
Quacky
$\sqrt{64}=8$.
So, using what I've said above, to what power must you raise 64 to get 8?
• Mar 8th 2010, 04:48 PM
minneola24
64 ^ .5 = 8

thanks

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

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

If $9^4=6561$

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

So it's easiest to work backwards.

$3^4=81$
So $81^?=3$
• Mar 8th 2010, 04:54 PM
Vertazontal
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 PM
minneola24
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 PM
Quacky
That's correct! (Happy)
• Mar 8th 2010, 05:15 PM
minneola24
Quote:

Originally Posted by Quacky
That's correct! (Happy)

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 PM
Quacky
$125^x=25^{x+1}$

I would do that by writing both as 5^{something}
$125=5^3$
$25=5^2$

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

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 PM
integral
Is this not a logarithm problem?
$3^x=27 \therefore log_3 27=x$
then

$x=\frac{log27}{log3}$
• Mar 8th 2010, 05:27 PM
Vertazontal
Quote:

Originally Posted by minneola24
awsome, thank you.

-snip-

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

for a i have

.5 = x

is that correct? thanks

A is x=2. 125 is 25^2, so 25^3 = 125^2

As for B, I have the answer but I can't explain why it works so I won't post it.
• Mar 8th 2010, 07:21 PM
Bacterius
Quote:

Originally Posted by integral
Is this not a logarithm problem?
$3^x=27 \therefore log_3 27=x$
then

$x=\frac{log27}{log3}$

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