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:
2. Do you mean $\displaystyle \frac{1}{9}x $ or $\displaystyle \frac{1}{9x}$ ?
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,
$\displaystyle \left(\frac{1}{3}\right)^{2x}=3^8$
$\displaystyle 3^{-2x}=3^8$
$\displaystyle -2x=8$
$\displaystyle x=-4$