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Solve the following equation: -2/3 ln (k-8) + 2/3 ln (k-1) = ln 4
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\[\frac{ -2 }{ 3 }\ln(k-8) + \frac{ 2 }{ 3 }\ln(k-1) = \ln4\]
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can use ... ln x + ln y = ln xy
Yes I know that but how would you solve that? (k-8)^-2/3 * (k-1)^2/3 = 4
can factor out 2/3 first...
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How?
2/3 ln [-(k-8)(k-1)] = ln 4 2/3 ln [(-k+8)(k-1)] = ln 4 \[\sqrt[3]{(-k+8)^2(k-1)^2}=4\]
Hmmm ok. But then if we're solving it don't we then get a quartic equation
\[(-k+8)(k-1)=\sqrt{4^3}\]
\[2/3(\ln(k-1)-\ln(k-8))=\ln4\] \[2/3\ln((k-1)/(k-8))=\ln4\]\[\ln((k-1)/(k-8))^2=\ln64\] \[((k-1)/(k-8))^2=64\] \[(k-1)/(k-8)=8\] \[k-1=8k-64\] \[63=7k\] \[k=9\]
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Oh I see now! I just wasn't too sure about the indice changes. Thanks to both of you!
np...
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