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Solving for x... Can anybody help? I think I did it right but want to make sure. Log2 base 2 ^ x-5 = 5
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Do you mean\[\log_22^{x-5}\]?
If so, that's\[\log_22^{x-5}=(x-5)\log_22=(x-5).1=x-5\]
Yes that is the equation. Solving for X.
what is X?
Well, you can't solve for x unless \[\log_22^{x-5}=(something)\]
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x=10
If there's no '=' sign in your question, it's an 'expression' that you need to simplify.
x=10 does that help?
Sorry - I didn't see the 5
\[x-5=5 \rightarrow x=10\]
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Yes It does . Thank you . That is what I got.
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