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Chapter 5:Indices and Logarithms Solve the equation. Double checking. @ganeshie8
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\[\log _{2}x+2=3\log _{x}2\]
\[\log _{2}x+2=3\frac{ 1 }{ \log _{2}x }\]
\[x+2=3\frac{ 1 }{ x }\]
\[x(x+2=3\frac{ 1 }{ x })\]
\[x^2+2x=3\]
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hey let it be some other variable so that we wont get confused : \(\log_2 x = t\)
okay
\[t^2+2t=3\]
looks good :)
\[t^2+3t-t-3=0\]
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\[t(t+3)-1(t+3)=0\]
\[t=1~and~t=-3\]
\[\log _{2}x=1\]\[\log _{2}x=-3\]
*
\[x=2^1,x=2^{-3}\]
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Am i correct? @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
It says that x=2 and x=1/8 @ganeshie8
Thnx @ganeshie8
yw!
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