The answer is;
54.6
But I used my TI-84 to get that answer and I can not remember how to do these.
51-2x = 0.25 51 * 5-2x = 0.25 5-2x = 0.05 -2x*ln(5) = ln(0.05) x = ln(0.05)/[-2*ln(5)] = 0.931
hanks to the limitations of the browser through which questions are posted, it is not clear what the question is but, here goes: If the question was ea = 35, then the answer is a = ln(35) where ln are the natural logarithms.
I assume the question is NOT about ln(a*b) = ln(a) + ln(b) because that is true for all positive real a and b. Instead, you want a solution to ln(a) * b = ln(a) + ln(b) or, ln(a) * (b-1) = ln(b) ln(a) = ln(b)/(b-1) ln(a) = ln[b1/(b-1)] Exponentiating, a = b1/(b-1) For any real number b > 1, a given by the above equation will meet your requirements.
ln(a) = 5.3 a = e5.3
18
To find ln 2.33, you need a calculator. It is the solution of the equation e^x = 2.33. ln 2.33 = 0.84586 (using a calculator)
3=lnx e^3=x
In the equation ln(x) = 5, the solution is x = (about) 148.4. To solve, simply raise e to the power of both sides and reduce... ln(x) = 5 eln(x) = e5 x = 148.4
-3 + ln(x) = 5 ln(x) = 8 eln(x) = e8 x = e8 x =~ 2981
51-2x = 0.25 51 * 5-2x = 0.25 5-2x = 0.05 -2x*ln(5) = ln(0.05) x = ln(0.05)/[-2*ln(5)] = 0.931
Pi^2 times the square root of 0.3 = Ln Xi for Ln Yi it's a little tougher. (0.3^4 times the value of Ln Xi)/17.2 Hope this helped. (not)
In the Clausius-Clapeyron equation, the variable "e" typically represents the base of the natural logarithm, which is approximately equal to 2.71828. This constant is important in relating the change in pressure to the change in temperature for a phase transition.
the natural log, ln, is the inverse of the exponential. so you can take the natural log of both sides of the equation and you get... ln(e^(x))=ln(.4634) ln(e^(x))=x because ln and e are inverses so we are left with x = ln(.4634) x = -0.769165
The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)
hanks to the limitations of the browser through which questions are posted, it is not clear what the question is but, here goes: If the question was ea = 35, then the answer is a = ln(35) where ln are the natural logarithms.
e = 2.71828183 (approximately)The definition of ln is this: ln x = y when e ^ y = x. It's an inverse property... So ln x means "find out what value y would need to have so that e ^ y equals x" Since e ^ 1 = e, ln e has to equal 1. because in line equation to signify that the task/job is done. This is why it is equal to 1.cause you add them and it just does
To be logarithmic the equation would be, ( otherwise just linear ) 3x + 7 = 20 subtract 7 from each side 3x = 13 now, you know the answer is between x = 2 and x = 3, so I use natural logs both sides ln(3x) = ln(13) as this is a logarithmic operation you can bring down the x in front of the ln sign on the left x ln(3) = ln(13) divideboth sides by ln(3) x = ln(13)/ln(3) ( not ln(13/3)!!!!! ) x = 2.334717519 -------------------------check in original equation 3(2.334717519) + 7 = 20 13 + 7 = 20 20 = 20 --------------checks