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1+4/3log little (x-3) 4 = 11/3
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what a pain
oh not really
ya i cant figure it out
1) subtract 1 from both sides get \[\frac{3}{4}\log_{x-3}(4)=\frac{8}{3}\]
rather \[\frac{4}{3}\log_{x-3}(4)=\frac{8}{3}\]
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multiply both sides by 4/3to get \[ \log_{x-3}(4)=\frac{8}{3}\times \frac{3}{4}=2\]
rewrite in equivalent exponential form to get \[(x-3)^2=4\] and the solve the quadratic for x
\[x-3=2\] \[x=5\]
you can forget about \[x-3=-2\] \[x=1\] because the base of a logarithm cannot be negative
thanks!
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