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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

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?

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

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

yes

that was a question

and the answer is 15

woudnt it be 17?

move to be 15

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

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