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\[\log_{21}(x-84)=3-\log_{21}x\] solve
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\[\log_{21}(x-84)=3-\log_{21}x \] \[\log_{21}(x-84)=\log_{21}21^3-\log_{21}x \] \[\log_{21}(x-84) =\log_{21}\frac{9261}{x} \] \[x-84=\frac{9261}{x}\] \[x^2-84x-9261=0\] \[(x-147)(x+63)=0\] \[x=147, -63\]
Discard -63 as it causes a negative argument
Thank you for writing this out! Which rule is used changing the "3?"
Well, actually I made it harder than it is. Try this: Add log base 21 of x to both sides and then use the first law of logs to write it as a product
\[\log_{21}(x-84)+\log_{21}x=3 \] \[\log_{21}(x-84)(x)=3 \] \[\log_{21}(x^2-84x)=3\] \[21^3=x^2-84x\] \[x^2-84x-9261=0\]
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\[Log _{a} a = 1 \]
There you go.
Thank you. :-)
yw
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