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Mech88
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2logx to the base 3  3log4 to the base 2 = log1 to the base b
 9 months ago
 9 months ago
Mech88 Group Title
2logx to the base 3  3log4 to the base 2 = log1 to the base b
 9 months ago
 9 months ago

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

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

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

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

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

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