condense the expression: 1/2Log5 16-3log5 x+4Log5y

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condense the expression: 1/2Log5 16-3log5 x+4Log5y

Mathematics
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first move the coefficients that are infront of the logarithms to the right
meaning log\[\log_{5} 16^{1/2}\]
ignore the first log my equations button is not working

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but the log expression below is correct use that example to simplify the other two
then let me know what you got
log5y^x+4?
\[\log_{5} 4-\log_{5} x ^{3}+\log_{5} y ^{4}\]
thankyou for helping me their just really hard for me ;(
no problem
have time for more?
we aren't done yet sorry, now when you are subtract two logs with the same base do you condense them by multiplying them together or dividing them?
dividing
yes so log base 4 is on top the numerator making the other two...?
on the bottom being multiplied together
idk
because they were being added
so can you show me what your answer would be?
4/x^3+y^4
when you are adding two logs together with the same base then you multiply them together
okay
\[\log_{5} (4/x ^{3}y ^{4})\]
does that make sense
yes
i just want to make sure you understand rather than just giving you the answer so that later you can ace a test on this if you need to
yah this good
ok why we try another problem. this will be the last one i can help with, then i have to go alright?
okay
solve the equation; 15+2 log2 x=31
so when solving for any variable the whole point is to get the variable by itself so what would be one thing we can move away from the side with x easily?
yes
that was a question
and the answer is 15
woudnt it be 17?
i didn't mean the final answer, i meant that it will be easy to move the fifteen since the fifteen is not attached to anything
move to be 15
yes we subtract fifteen from both sides then we simplify the logarithm but first what does the new equation look like?
it will equeal 16instead of 31
yes
know we focus on the log
;)
okay!
what can we move to condense the logarithm more?
the 2 in front
yes
ah yes!!
so what do we have now?
logx=16
no... \[\log_{2} x ^{2}=16\]
does that make sense?
yes
okay a trick that i learned is you put the same base of the log you want to remove under both sides of the equation like so: \[2^{\log_{2}x ^{2} }=2^{16}\]
now what can you simplify?
exponents
yes \[x ^{2}=2^{16}\] is our new equation as 2^log base 2 equals one
now what would be the next step?
logx=4
why would we do that when we already have x by itself. what we would do is simplify 2^16=65536
then x^2=65536
65536?
yes now what would you do to get x all by itself?
take the root of x and the other side
x=256
does that make sense?
thats what i thought
\[\sqrt{x ^{2}}=\sqrt{65536}=x=256\]
omg thankyou so much
no problem I have to go now, but i wish you the best of luck :)
thankyou

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