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How would you solve ln 4 plus 3 ln x equals 5 ln 2?

Ln 4 + 3Ln x = 5Ln 2 Ln 4 + Ln x3= Ln 25 = Ln 32 Ln x3= Ln 32 - Ln 4 = Ln (32/4) = Ln 8= Ln 2


What is equivalent to ln 6 plus ln 4?

You can also write this as ln(6 times 4)


How do you solve for x 3 ln x - ln 3x equals 0?

3 ln(x) = ln(3x)ln(x3) = ln(3x)x3 = 3xx2 = 3x = sqrt(3)x = 1.732 (rounded)


How do you write ln a equals 5.3 in exponential form?

ln(a) = 5.3 a = e5.3


Solve or x 2 ln 9 plus 2 ln 5 equals 2 ln x minus 3?

2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)


What is the answer to 2 minus Ln times 3 minus x equal 0?

so, if 2 minus Ln times 3 minus x equals 0, then 2 minus Ln times 3 equals x, therefore 2 minus Ln equals x divided by three, so Ln + X/3 = 2 therefore, (Ln + [X/3]) = 1


LM equals 4 in LN equals?

4.60


Is there a function where the first derivative goes to infinity for x going to 0 and where the first derivative equals 0 when x is 1?

Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.


What is the derivative of y equals xlnx?

Use the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln x


Mn equals 72 ln equals?

14\2


Do any positive real numbers alpha and beta exist such that lnalpha times beta equals lnalpha plus lnbeta and if so what are they?

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.


How do you solve ln 5 plus ln x equals 1?

5 + (-4) = 1 x = -4