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Assuming you mean 'logarithm to the base 'e' ( natural logarithm.
On the calculator its symbol is 'ln'.
Hence ;ln 2 = 0.69314718....
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How do you differentiate x to the exponent ln2?
In (x^(ln2)), ln2 is a constant, so the Power Rule can be used. d/dx (x^ln2)=(ln2)x^(ln2-1)
Lnx plus Ln2 equals 2 plus lnx?
NO! Lnx + Ln2= 2 + Lnx implies Ln2 = 2 which implies 2 = e2 which is simply not true.
What is the derivative of the square root of 2 raised to the power of cos x?
The derivative of 2^x is 2^x * ln2 so the derivative of 2^cosx * ln2 multiplied by d/dx of cox, which is -sinx so the derivative of the inside function is -sinx * 2^cosx *ln2. As to the final question, using the chain rule, d/dx (2^cosx)^0.5 will equal half of (2^cosx)^-0.5 * -sinx * 2^cosx * ln2
What is In10 plus Inx-In2 as one logarithmic?
12
A sample of a radioactive substance decayed to 95 percent of its original amount after a year.what is the half life of the substance?
The half-life of a radioactive nuclide when 95% of it is left after one year is 13.5 years. AT = A0 2(-T/H) 0.95 = (1) 2(-1/H) ln2(0.95) = -1/H H = -1/ln2(0.95) H = 13.5
