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anonymous
 5 years ago
Solving for x... Can anybody help? I think I did it right but want to make sure.
Log2 base 2 ^ x5 = 5
anonymous
 5 years ago
Solving for x... Can anybody help? I think I did it right but want to make sure. Log2 base 2 ^ x5 = 5

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you mean\[\log_22^{x5}\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If so, that's\[\log_22^{x5}=(x5)\log_22=(x5).1=x5\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes that is the equation. Solving for X.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, you can't solve for x unless \[\log_22^{x5}=(something)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If there's no '=' sign in your question, it's an 'expression' that you need to simplify.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry  I didn't see the 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x5=5 \rightarrow x=10\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes It does . Thank you . That is what I got.
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