1. ## 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

2. 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 $\displaystyle log_b(a^k) = k \times log_b(a)$ ($\displaystyle b >0 , \neq 1$) 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 $\displaystyle 2^{8x} = 2^2$

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

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

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

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

2. Do you mean $\displaystyle \frac{1}{9}x$ or $\displaystyle \frac{1}{9x}$ ?

3. 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,

$\displaystyle 2^8=256$

(1) $\displaystyle 256^x=4$

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

$\displaystyle 8x=2$

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

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

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

Then,

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

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

$\displaystyle -2x=8$

$\displaystyle x=-4$

4. (1/9)x=6561

It doesn't have the x as an exponent just as a multiplier

5. Originally Posted by leilani
It doesn't have the x as an exponent just as a multiplier
Like this $\displaystyle \frac{1}{9}x=6561$??

This one is easier than the other one, then. Just multiply both sides by 9.

6. 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

7. 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.

8. 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