# exponentials....i need help

• Apr 23rd 2009, 09:23 AM
leilani
exponentials....i need help
I have two problems I am stuck on. I missed lecture and need help with both
1. 256^x=4

2. 1/9x=6561

Thanks to anyone who can help me. Leilani
• Apr 23rd 2009, 09:39 AM
e^(i*pi)
Quote:

Originally Posted by leilani
I have two problems I am stuck on. I missed lecture and need help with both
1. 256^x=4

2. 1/9x=6561

Thanks to anyone who can help me. Leilani

1. Take the log of both sides (you can pick any base you like)

You can use the rule $log_b(a^k) = k \times log_b(a)$ ( $b >0 , \neq 1$) to bring x to the front and remove it as a power:

As 256 = 2^8 and 4 = 2^2 we can write the equation as $2^{8x} = 2^2$

Spoiler:
Take logs of both sides (I will use base e):

$8x \times ln(2) = 2ln(2)$

$x = \frac{1}{4}$

(note that using base 2 in your logarithm will get $log_2(2) = 1$ so they get cancelled a step earlier)

2. Do you mean $\frac{1}{9}x$ or $\frac{1}{9x}$ ?
• Apr 23rd 2009, 09:50 AM
masters
Quote:

Originally Posted by leilani
I have two problems I am stuck on. I missed lecture and need help with both
1. 256^x=4

2. 1/9x=6561

Thanks to anyone who can help me. Leilani

Hi leilani,

$2^8=256$

(1) $256^x=4$

$2^{8x}=2^2$

$8x=2$

$x=\frac{1}{4}$

(2) Did you mean to write this: $\left(\frac{1}{9}\right)^x=6561$? I can't see your picture for some reason.

If so, know that $3^8=6561$ and $\frac{1}{9}=\left(\frac{1}{3}\right)^2=3^{-2}$

Then,

$\left(\frac{1}{3}\right)^{2x}=3^8$

$3^{-2x}=3^8$

$-2x=8$

$x=-4$
• Apr 23rd 2009, 10:09 AM
leilani
(1/9)x=6561

It doesn't have the x as an exponent just as a multiplier
• Apr 23rd 2009, 10:21 AM
masters
Quote:

Originally Posted by leilani
It doesn't have the x as an exponent just as a multiplier

Like this $\frac{1}{9}x=6561$??

This one is easier than the other one, then. Just multiply both sides by 9.
• Apr 23rd 2009, 10:23 AM
leilani
Thanks, I got it. I set it up just like the other one. I think there was a mistake how they wrote the problem. When I worked it out just like the other one I got the right answer. Thanks very much! Leilani
• Apr 23rd 2009, 10:24 AM
masters
Quote:

Originally Posted by leilani
Thanks, I got it. I set it up just like the other one. I think there was a mistake how they wrote the problem. When I worked it out just like the other one I got the right answer. Thanks very much! Leilani

What was the right answer since we don't know how it was meant to be written in the first place?? Just for yuks.
• Apr 23rd 2009, 10:53 AM
leilani
Quote:

Originally Posted by masters
What was the right answer since we don't know how it was meant to be written in the first place?? Just for yuks.

X=-4
Thanks for the help