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solve for x. log[4](x+4)+log[4](x+16)=3 a. 16 b. 0 or -20 c. 0 d. -20 e. none of the above
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c, 1+2=3
\[\log _{4}(0+4)=1\]\[\log _{4}(0+16)=2\]
Need to check that -20 out also
nice
Need to check that -20 out also\[\log _{4}(-20+4)=\log _{4}-16\] don't think that is a 2. Anyone want to comment.
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\[4^{x}=-16\] ?????
4*4*(-1)
is only 0
or just
O.K Thanks
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nice
I guess that wraps it up Yardbird84
thank you for your help
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