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Mathematics 19 Online
OpenStudy (anonymous):

Solve 2^(2x+1)=20

OpenStudy (anonymous):

try taking the ln of log of both sides

OpenStudy (anonymous):

Should I simplify it first?

OpenStudy (anonymous):

There is nothing to simplify before taking the log

OpenStudy (anonymous):

I used the quotient law then I am stuck

OpenStudy (anonymous):

what did you get when you log both sides?

OpenStudy (unklerhaukus):

2^(2x+1) = 20 yes simplify first, cancel a factor of 2 2^(2x) = 10 then use an index law on the lefthand side (2^2)^x = 10

OpenStudy (unklerhaukus):

when you have b^x = 10 take the log of both sides (to base b) x = log_b (10) = ln 10 / ln b (change of base formula) and use another log law to simplify this

OpenStudy (anonymous):

\[2^{2x+1}=10\]\[(2x+1)\ln 2=\ln 10\]\[2x+1=\frac{ \ln10 }{\ln2 }\]\[x=\frac{ 1 }{ 2 }(\frac{ \ln10 }{\ln2 }-1)\]

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