1. ## [SOLVED] logs

Does anyone understand these two problems:

Problem 1:
(hint: )

Problem 2:

for 1, I got -2k

& for 2, I got 3

which were wrong

2. Originally Posted by lsnyder
Does anyone understand these two problems:

Problem 1:
(hint: )
Incorporating the hint, this is equivalent to:

$\displaystyle \log_{1048576}\left[\left(\left[1048576^{\frac{1}{10}}\right]^{-2}\right)^k\right]=\log_{1048576}\left[\left(1048576\right)^{-\frac{2k}{10}}\right]$ $\displaystyle =-\frac{k}{5}\log_{1048576}\left(1048576\right)=\col or{red}\boxed{-\frac{k}{5}}$

Problem 2:

for 1, I got -2k

& for 2, I got 3

which were wrong
Note that $\displaystyle 3^{6\log_3 5-5\log_3 6}=3^{6\log_3 5}3^{-5\log_3 6}=3^{\log_3\left(5^6\right)}3^{\log_3\left(6^{-5}\right)}$

Can you finish the simplification?

Does this make sense?

3. okay i somewhat ubderstand 1.
i just don't understand where u got 1/10.
other than that i get it.

& on problem 2,
i am very lost.
where did the 3 come from in between 5 and -5

4. oh nevermind, i figured out the 1/10 in the 1st problem

but i still don't follow the second one

5. Originally Posted by lsnyder
okay i somewhat ubderstand 1.
i just don't understand where u got 1/10.
other than that i get it.
In the hint, $\displaystyle 1048576=4^{10}\implies 1048576^{\frac{1}{10}}=4$. Thus, in your expression, I substituted $\displaystyle 1048576^{\frac{1}{10}}$ for $\displaystyle 4$.

Does this clarify things?

& on problem 2,
i am very lost.
where did the 3 come from in between 5 and -5
I applied rules of exponents:

$\displaystyle a^{b-c}=a^ba^{-c}$

6. yea i got the 1/10 after i posted i didn't understand.
but problem 1 i understand.

now on problem 2,
i see what you did with the 3.
i have 1 question though.
Will all the threes cancel out?
or is there something else i am missing?

7. nevermind i got problem 2

if any1 cares the answer was:

15625/7776