Question

Help needed with logs... equations to come

Answer 1

#1 Simplify:\[5 \log _{2}k-8 \log _{2}m+10 \log _{2}n\]
#2 Solve: \[2 \log _{2} 2+2 \log _{2} 6-\log _{2}3x=3\]

Answer 2

\(\bf \begin{array}{llll}
5 log _2k-8 log _2m+10 log _2n\\
\textit{recall that }log_ab^x=xlog_ab\\\\ \quad\
\bf [5 log _2k-8 log _2m]+10 log _2n\\
\textit{recall that }log(a)-log(b)=log\left(\frac{a}{b}\right)\\\\ \quad \
\bf [5 log _2k-8 log _2m]+[10 log _2n]\\
\textit{recall that }log(a)+log(b)=log(a\cdot b)
\end{array}\)

Answer 3

do you understand what jdoe0001 is telling you?

Answer 4

Not really

Answer 5

the first step\[\log_a b^x=x\cdot \log_a b\]reminds you that you can move the number in front of the logs into the exponent of what you're taking the log of, for example
\[5 \log_2 k=\log_2 k^5\]so the first step would be to do that for all three terms

Answer 6

The answers for each are #1:
\[7\log _{2}(k-m+n)\]
\[7\log _{2}\frac{ kn }{ m }\]
\[\log _{2}\frac{ 50kn }{ 8m }\]
\[\log _{2}\frac{ k^5n ^{10} }{ m^8 }\]

#2
x=2
x=6
x=16
x=18

Answer 7

give me a minute to go through the question myself and check your work

Answer 8

oh those are multiple choice answers. do the first step for #1, move the exponents up. what do you get?

Answer 9

I got the answer on the first one \[\log _{2}\frac{ k^5n ^{10} }{ m^8 }\]

Answer 10

that's correct for #1

Answer 11

hmmm http://www.chilimath.com/algebra/advanced/log/images/466x428xrules,P20of,P20exponents.gif.pagespeed.ic.1z7-Ekl5L5.png

Answer 12

I meant, use rule 3 listed there, then rule 2, then rule 1

Answer 13

I got 12 for the 2nd one, but that isn't an answer

Answer 14

yeah 12 isn't correct

Answer 15

I've got to go eat dinner... will come back to it afterwards

Answer 16

\(\bf log_22+log_26-log_2(3x)=3\\ \quad \\ \quad \\
log_2(2\cdot 6)-log_2(3x)=3\implies log_2\left[\cfrac{(2\cdot 6)}{(3x)}\right]=3\\ \quad \\
\textit{log cancellation rule of }a^{log_ax}=x\\ \quad \\
log_2\left[\frac{(2\cdot 6)}{(3x)}\right]=3\implies {\Large 2^{log_2\left[\frac{(2\cdot 6)}{(3x)}\right]}=2^3}\implies \cfrac{(2\cdot 6)}{(3x)}=2^3\)

Answer 17

The first term is pretty easy to simplify\[2\log_2 2=\log_2 2^2=\log_2 4=2\]so we have\[2+2\log_2 6-\log_2 3x=3 \]moving the terms that don't have a variable to the right,\[-\log_2 3x=3-2\log_2 6-2\]cleaning up a bit,\[\log_2 3x=2\log_2 6-1\]moving the 2 in 2log_2(6) into the exponent,\[\log_2 3x=\log_2 6^2-1\]the 1 is a problem, but\[1=log_2 2\]so we have\[\log_2 3x=\log_2 6^2-\log_2 2\]using the laws of logarithms from the last question,\[\log_2 3x=\log_2 \frac{6^2}{2}\]\[3x=\frac{6^2}{2}=\frac{36}{2}=18\]\[x=\frac{18}{3}=6\]

Answer 18

@jdoe0001 copied the question wrong, but the method is basically the same as I did, just completed in a different order. Just keep in mind their final answer won't be correct.

Answer 19

Thank you @jdoe0001 and @Xeph
I got it right, but your way is easier to understand.