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2logx to the base 3 - 3log4 to the base 2 = log1 to the base b

Algebra
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What is the question?
Oh the b?
Solve for x

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Other answers:

is this the question? \[2\log_{3}(x) - 3\log_{2}(4) = \log_{b}(1)\]
Yes... That's the question thank you
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...
How would I do that? I also get to that step but can't get further
ok... so this is how I would do the question, after re-reading 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
Thank you so much

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