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Mech88
 2 years ago
2logx to the base 3  3log4 to the base 2 = log1 to the base b
Mech88
 2 years ago
2logx to the base 3  3log4 to the base 2 = log1 to the base b

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SolomonZelman
 2 years ago
Best ResponseYou've already chosen the best response.0What is the question?

campbell_st
 2 years ago
Best ResponseYou've already chosen the best response.1is this the question? \[2\log_{3}(x)  3\log_{2}(4) = \log_{b}(1)\]

Mech88
 2 years ago
Best ResponseYou've already chosen the best response.0Yes... That's the question thank you

campbell_st
 2 years ago
Best ResponseYou've already chosen the best response.1ok... so \[\log_{2}(4) = \log_{2}(2^2) = 2\] so you have \[\log_{3}(x^2)  6 = \log_{b}(1)\] so I'd suggest using change of base...

Mech88
 2 years ago
Best ResponseYou've already chosen the best response.0How would I do that? I also get to that step but can't get further

campbell_st
 2 years ago
Best ResponseYou've already chosen the best response.1ok... so this is how I would do the question, after rereading it the log of 1 to any base is always zero... its just a fact. and you saw how I managed the base 2 log of 4 so you have \[2\log_{3}(x)  6 = 0\] which becomes \[2 \log_{3}(x) = 6\] divide both sides of the equation by 2 \[\log_{3}(x) = 3\] now raise each side of the equation to the power of 3 \[3^{\log_{3}(x)} = 3^3\] which is x = 27
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