log(3x+5)-log(x-5)=log(8)

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log(3x+5)-log(x-5)=log(8)

Mathematics
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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.

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i got it so log(3x+5/x-5)=log8 and dont know where to go from there
hint: \[\huge\rm log_b x = \log_b y\] \[\huge\rm\cancel { log_b} x = \cancel{\log_b} y\] x=y
so 3x-5/x-5=8

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Other answers:

yep now simple algebra
i got -20 but the answer is 9
how did you get -20?
i cross multiplied
so 8x-40=3x+5
solve for x
i got -20
show ur work
oh i found my mistake thank you!
r u sure ? u got it ?
yep! can i ask you one more though?
alright
log8+3logx=3 which i know is log8+logx^3=3 but what do you do if theres no log on the other side?
apply log properties which one you should apply ?
change of base?
yeah or convert log to exponential form just like we did before
nope there is a variable so you can't use change of base formula
how would you do that though because isnt it 8x^3=3
log (8x^3) = 3
\[\log = \log_{10}\]
but idk how you would make that exponential
|dw:1432768266869:dw|
but theres no base
there is i just posted that
\[\log = \log_{10}\]
oh ok so then i get 10 though and the answe r is 5
xo x = 5 ?
10^3=x^3 which is 10
oh i forgot the 8
oh ok so x=5 got it thanks
yep

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