# how do you solve 3log2x=4? please explain also..?

• 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

• 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.

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

• 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.

• 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)

• sorry, you didn't write the question in a specific way.

is it log base 2x? log base 2? log base 10?

Sorry

• When writing an equation, please be precise, especially with log function. Is it base 2 or base e?