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I am assuming that your question is about log base 3 of (x plus 1) plus log base 2 of (x-1).
{log[(x + 1)^log2} + {log[(x - 1)^log3}/log(3^log2) where all the logs are to the same base - whichever you want. The denominator can also be written as log(3^log2)
This can be simplified (?) to log{[(x + 1)^log2*(x - 1)^log3}/log(3^log2).
As mentioned above, the expression can be to any base and so the expression becomes
in base 2: log{[(x + 1)*(x - 1)^log3}/log(3) and
in base 3: log{[(x + 1)^log2*(x - 1)}/log(2)
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Log x plus 7 equals 1 plus Log x-1?
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Log(3 * 1/3) = log(1) = 0
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What is X3 plus X2 plus X plus 1 divided by X 1?
2x2+7/x1
X minus 1 plus 2 plus log x equals 3. find x?
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acording to me the value is 0 because the value of log 1 at any base is always 0.
What is the answer to Log x plus 4 minus log6 equals 1?
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Suppose aneq 0. Compute log 2a 2b in terms of a and b.?
Due to the rubbish browser that we are compelled to use, it is not possible to use any super or subscripts so here goes, with things spelled out in detail: log to base 2a of 2b = log to base a of 2b/log to base a of 2a = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + (log to base a of a)] = [(log to base a of 2) + (log to base a of b)] / [(log to base a of 2) + 1]
Solve ln y equals xln e?
ln is the natural logarithm. That is it is defined as log base e. As we all know from school, log base 10 of 10 = 1 just as log base 3 of 3 = 1, so, likewise, log base e of e = 1 and 1.x = x. so we have ln y = x. Relace ln with log base e, and you should get y = ex
Log x plus 7 equals 1 plus Log x-1?
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What is log to the base of 2 0.5?
- 1
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It is zero
How do you swap two numbers without third variables?
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Differentiate log x?
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