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

- anonymous

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

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- anonymous

first move the coefficients that are infront of the logarithms to the right

- anonymous

meaning log\[\log_{5} 16^{1/2}\]

- anonymous

ignore the first log my equations button is not working

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- anonymous

but the log expression below is correct use that example to simplify the other two

- anonymous

then let me know what you got

- anonymous

log5y^x+4?

- anonymous

\[\log_{5} 4-\log_{5} x ^{3}+\log_{5} y ^{4}\]

- anonymous

thankyou for helping me their just really hard for me ;(

- anonymous

no problem

- anonymous

have time for more?

- 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

dividing

- anonymous

yes so log base 4 is on top the numerator making the other two...?

- anonymous

on the bottom being multiplied together

- anonymous

idk

- anonymous

because they were being added

- anonymous

so can you show me what your answer would be?

- anonymous

4/x^3+y^4

- anonymous

when you are adding two logs together with the same base then you multiply them together

- anonymous

okay

- anonymous

\[\log_{5} (4/x ^{3}y ^{4})\]

- anonymous

does that make sense

- anonymous

yes

- 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

yah this good

- anonymous

ok why we try another problem. this will be the last one i can help with, then i have to go alright?

- anonymous

okay

- anonymous

solve the equation; 15+2 log2 x=31

- 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

yes

- anonymous

that was a question

- anonymous

and the answer is 15

- anonymous

woudnt it be 17?

- 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

move to be 15

- anonymous

yes we subtract fifteen from both sides then we simplify the logarithm but first what does the new equation look like?

- anonymous

it will equeal 16instead of 31

- anonymous

yes

- anonymous

know we focus on the log

- anonymous

;)

- anonymous

okay!

- anonymous

what can we move to condense the logarithm more?

- anonymous

the 2 in front

- anonymous

yes

- anonymous

ah yes!!

- anonymous

so what do we have now?

- anonymous

logx=16

- anonymous

no... \[\log_{2} x ^{2}=16\]

- anonymous

does that make sense?

- anonymous

yes

- 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

now what can you simplify?

- anonymous

exponents

- anonymous

yes \[x ^{2}=2^{16}\] is our new equation as 2^log base 2 equals one

- anonymous

now what would be the next step?

- anonymous

logx=4

- anonymous

why would we do that when we already have x by itself.
what we would do is simplify 2^16=65536

- anonymous

then x^2=65536

- anonymous

65536?

- anonymous

yes now what would you do to get x all by itself?

- anonymous

take the root of x and the other side

- anonymous

x=256

- anonymous

does that make sense?

- anonymous

thats what i thought

- anonymous

\[\sqrt{x ^{2}}=\sqrt{65536}=x=256\]

- anonymous

omg thankyou so much

- anonymous

no problem I have to go now, but i wish you the best of luck :)

- anonymous

thankyou

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