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Does anyone want to help me out with a few log questions?
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What's the question?
\[(4/7)^x=6^1-x \] @Ankh
6 to the power of 1?
thats what it says lol but solve for x
oh wait its 6^(1-x)
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Well I reached a simpler form, but not sure how to continue from there, it's quite a challenging problem.
Oh wait I think i solved it.
okay thanks, i just needed to find it in decimal form bc its multiple choice
did u get..\[\ln6/\ln(4/7)+\ln6\]
\[\left( \frac{ 4 }{ 7 } \right)^x=6^{x-1}\] \[\frac{ 4^x }{ 7^x } = 6^1 \times 6^{-x}\] \[\frac{ 4^x }{ 7^x } = \frac{ 6 }{ 6^x }\] \[4^x \times 6^x = 7^x \times 6\] \[\frac{ 24^x }{ 7^x } = 6\] \[\log_{24/7} (6)=x\]
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okay thank you, i get it now!
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