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Mech88
Group Title
2logx to the base 3  3log4 to the base 2 = log1 to the base b
 one year ago
 one year ago
Mech88 Group Title
2logx to the base 3  3log4 to the base 2 = log1 to the base b
 one year ago
 one year ago

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SolomonZelman Group TitleBest ResponseYou've already chosen the best response.0
What is the question?
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.0
Oh the b?
 one year ago

Mech88 Group TitleBest ResponseYou've already chosen the best response.0
Solve for x
 one year ago

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

Mech88 Group TitleBest ResponseYou've already chosen the best response.0
Yes... That's the question thank you
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
ok... 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...
 one year ago

Mech88 Group TitleBest ResponseYou've already chosen the best response.0
How would I do that? I also get to that step but can't get further
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
ok... 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
 one year ago

Mech88 Group TitleBest ResponseYou've already chosen the best response.0
Thank you so much
 one year ago
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