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Mathematics 19 Online
OpenStudy (anonymous):

Chapter 5:Indices and Logarithms Solve the equation. Double checking. @ganeshie8

OpenStudy (anonymous):

\[\log _{2}x+2=3\log _{x}2\]

OpenStudy (anonymous):

\[\log _{2}x+2=3\frac{ 1 }{ \log _{2}x }\]

OpenStudy (anonymous):

\[x+2=3\frac{ 1 }{ x }\]

OpenStudy (anonymous):

\[x(x+2=3\frac{ 1 }{ x })\]

OpenStudy (anonymous):

\[x^2+2x=3\]

ganeshie8 (ganeshie8):

hey let it be some other variable so that we wont get confused : \(\log_2 x = t\)

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

\[t^2+2t=3\]

ganeshie8 (ganeshie8):

looks good :)

OpenStudy (anonymous):

\[t^2+3t-t-3=0\]

OpenStudy (anonymous):

\[t(t+3)-1(t+3)=0\]

OpenStudy (anonymous):

\[t=1~and~t=-3\]

OpenStudy (anonymous):

\[\log _{2}x=1\]\[\log _{2}x=-3\]

OpenStudy (jhannybean):

*

OpenStudy (anonymous):

\[x=2^1,x=2^{-3}\]

OpenStudy (anonymous):

Am i correct? @ganeshie8

ganeshie8 (ganeshie8):

Excellent work ! lets triple check wid wolf http://www.wolframalpha.com/input/?i=solve+%5Clog_2+x%2B2%3D3%5Clog_x+2+over+reals

OpenStudy (anonymous):

It says that x=2 and x=1/8 @ganeshie8

OpenStudy (anonymous):

Thnx @ganeshie8

ganeshie8 (ganeshie8):

yw!

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