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

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- Apr 23rd 2009, 09:23 AMleilaniexponentials....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 AMe^(i*pi)
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__:

2. Do you mean $\displaystyle \frac{1}{9}x $ or $\displaystyle \frac{1}{9x}$ ? - Apr 23rd 2009, 09:50 AMmasters
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,

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

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

$\displaystyle -2x=8$

$\displaystyle x=-4$ - Apr 23rd 2009, 10:09 AMleilani
(1/9)x=6561

It doesn't have the x as an exponent just as a multiplier - Apr 23rd 2009, 10:21 AMmasters
- Apr 23rd 2009, 10:23 AMleilani
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 AMmasters
- Apr 23rd 2009, 10:53 AMleilani