1. ## Backward Exponets

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

2. Hint:
$\sqrt x = x^\frac{1}{2}$

3. still confused.

4. $\sqrt{64}=8$.
So, using what I've said above, to what power must you raise 64 to get 8?

5. 64 ^ .5 = 8

thanks

i cant do 64 ^ .5 to get 2 so what do I do?

6. 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 $81^?=3$

7. 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

8. so to get 81 to 3 you do 81^.25

hjowever for 64 -> 2 how do i do 2^1/6 to check?

9. That's correct!

10. Originally Posted by Quacky
That's correct!
awsome, thank you.

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

for a i have

.5 = x

is that correct? thanks

11. $125^x=25^{x+1}$

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

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

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

13. 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.

14. 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 (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.