7 Answers
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Inverse of a log is 10^x so therefore we can solve using a bit of algebra:
3log2x=4
log2x=4/3 -- Divide by 3 on both sides
2x=10^(4/3) -- Inverse of log is 10^x. Apply to both sides
x={10^(4/3)}/2 -- Divide both sides by 2
Now solve.
x = 10.77217345 or about that.
We can sub in for X to see that is correct.
3log2{10.77217345}=4
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Hi. I just trimmed your question to "How do you solve 3log2x=4?" and entered it in to my browser and this was the result. It apparently had been asked previously and Yahoo! Answers.
http://answers.yahoo.com/question/index?qid=200904...
Solve 3log2x = 4. round to the nearest ten-thousanth.? - Yahoo ...
Apr 11, 2009 ... Block User. Best Answer - Chosen by Voters ...
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Assuming "log" is the base 10 log...
3 log 2x = 4
log 2x = 4/3
10^ (log 2x) = 10^ (4/3)
2x = 10 ^ (4/3)
x = (10 ^ (4/3)) / 2
x = one half the cube root of 10000
If it's the natural log (base e) then the answer is
(e ^ (4/3)) /2
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it depends on if you mean 3log(2x) = 4, or 3log(base2)x = 4. in either case, the first step is to divide by 3. if you are talking about (natural) log(2x), you would then exponentiate (by e) to obtain:
2x = e^(4/3), so x=(e^(4/3))/2. if however, you are talking about log(base2)x, you would have:
x = 2^(4/3), or twice the cube root of 2.
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2^3log2x=2^4
2^log2x^3=2^4
x^3=2^4
x^3=2*2^3
x=2*2^(1/3)
x=2^(4/3)
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sorry, you didn't write the question in a specific way.
is it log base 2x? log base 2? log base 10?
Sorry
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When writing an equation, please be precise, especially with log function. Is it base 2 or base e?
