anonymous
  • anonymous
condense the expression: 1/2Log5 16-3log5 x+4Log5y
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
first move the coefficients that are infront of the logarithms to the right
anonymous
  • anonymous
meaning log\[\log_{5} 16^{1/2}\]
anonymous
  • anonymous
ignore the first log my equations button is not working

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
but the log expression below is correct use that example to simplify the other two
anonymous
  • anonymous
then let me know what you got
anonymous
  • anonymous
log5y^x+4?
anonymous
  • anonymous
\[\log_{5} 4-\log_{5} x ^{3}+\log_{5} y ^{4}\]
anonymous
  • anonymous
thankyou for helping me their just really hard for me ;(
anonymous
  • anonymous
no problem
anonymous
  • anonymous
have time for more?
anonymous
  • anonymous
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?
anonymous
  • anonymous
dividing
anonymous
  • anonymous
yes so log base 4 is on top the numerator making the other two...?
anonymous
  • anonymous
on the bottom being multiplied together
anonymous
  • anonymous
idk
anonymous
  • anonymous
because they were being added
anonymous
  • anonymous
so can you show me what your answer would be?
anonymous
  • anonymous
4/x^3+y^4
anonymous
  • anonymous
when you are adding two logs together with the same base then you multiply them together
anonymous
  • anonymous
okay
anonymous
  • anonymous
\[\log_{5} (4/x ^{3}y ^{4})\]
anonymous
  • anonymous
does that make sense
anonymous
  • anonymous
yes
anonymous
  • anonymous
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
anonymous
  • anonymous
yah this good
anonymous
  • anonymous
ok why we try another problem. this will be the last one i can help with, then i have to go alright?
anonymous
  • anonymous
okay
anonymous
  • anonymous
solve the equation; 15+2 log2 x=31
anonymous
  • anonymous
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?
anonymous
  • anonymous
yes
anonymous
  • anonymous
that was a question
anonymous
  • anonymous
and the answer is 15
anonymous
  • anonymous
woudnt it be 17?
anonymous
  • anonymous
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
anonymous
  • anonymous
move to be 15
anonymous
  • anonymous
yes we subtract fifteen from both sides then we simplify the logarithm but first what does the new equation look like?
anonymous
  • anonymous
it will equeal 16instead of 31
anonymous
  • anonymous
yes
anonymous
  • anonymous
know we focus on the log
anonymous
  • anonymous
;)
anonymous
  • anonymous
okay!
anonymous
  • anonymous
what can we move to condense the logarithm more?
anonymous
  • anonymous
the 2 in front
anonymous
  • anonymous
yes
anonymous
  • anonymous
ah yes!!
anonymous
  • anonymous
so what do we have now?
anonymous
  • anonymous
logx=16
anonymous
  • anonymous
no... \[\log_{2} x ^{2}=16\]
anonymous
  • anonymous
does that make sense?
anonymous
  • anonymous
yes
anonymous
  • anonymous
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: \[2^{\log_{2}x ^{2} }=2^{16}\]
anonymous
  • anonymous
now what can you simplify?
anonymous
  • anonymous
exponents
anonymous
  • anonymous
yes \[x ^{2}=2^{16}\] is our new equation as 2^log base 2 equals one
anonymous
  • anonymous
now what would be the next step?
anonymous
  • anonymous
logx=4
anonymous
  • anonymous
why would we do that when we already have x by itself. what we would do is simplify 2^16=65536
anonymous
  • anonymous
then x^2=65536
anonymous
  • anonymous
65536?
anonymous
  • anonymous
yes now what would you do to get x all by itself?
anonymous
  • anonymous
take the root of x and the other side
anonymous
  • anonymous
x=256
anonymous
  • anonymous
does that make sense?
anonymous
  • anonymous
thats what i thought
anonymous
  • anonymous
\[\sqrt{x ^{2}}=\sqrt{65536}=x=256\]
anonymous
  • anonymous
omg thankyou so much
anonymous
  • anonymous
no problem I have to go now, but i wish you the best of luck :)
anonymous
  • anonymous
thankyou

Looking for something else?

Not the answer you are looking for? Search for more explanations.