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Mathematics 15 Online
OpenStudy (anonymous):

log2(5x)-log2(5)=100

OpenStudy (anonymous):

\[\log_{2} 5x-\log_{2}5=100\]

OpenStudy (anonymous):

log2(x) = 100

OpenStudy (anonymous):

Would I just do 5x-5=100?

OpenStudy (anonymous):

x= 2^100

OpenStudy (anonymous):

How? I don't understand.

OpenStudy (anonymous):

log 2(x) = 100 => x = 2^100

OpenStudy (anonymous):

How do I get from log2(5x)-log2(5)=100 to log 2(x)?

OpenStudy (anonymous):

5x/5 = x

hero (hero):

@petewe, you didn't explain anything that would help @melbel understand

OpenStudy (anonymous):

i'd assumed basic logarithm properties were known.

OpenStudy (anonymous):

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