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Q: Why does ln x plus 2 equal 2 plus ln x?

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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)

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

Do you mean ln(x-2), or ln(x)-2? If it is ln(x-2): 1/(x-2) If it is ln(x)-2: 1/x

ln(x+14)-lnx=3ln2 ln[(x+14)/x]=ln8 (x+14)/x=8 x+14=8x 14=7x 2=x x=2 Check this answer by plugging x=2 back into the original equation: ln(2+14)-ln(2)=3ln2 ln(16)-ln(2)=3ln2 ln(16/2)=3ln2 ln8=3ln2 3ln2=3ln2 There you go!

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

The order of operations is not quite clear here.If you mean (ln 2) + x, the derivate is 0 + 1 = 1.If you mean ln(2+x), by the chain rule, you get (1/x) times (0+1) = 1/x.

x^(ln(2)/ln(x)-1)

2x = 5 then x*ln(2) = ln(5) so that x = ln(5)/ln(2) = 2.3219 approx.

log3(x) + log9(x^3)=0 for any t >0 logt(x) = ln(x)/ln(t) so log9(x) = ln(x) / ln (9) = ln (x) / ( 2 * ln 3) = log3(x) /2 or log3(x) = 2 log9(x) log9(x^n) = n * log9(x) So log3(x) + log9(x^3) = log3(x) + 6 log3(x) = 7 log3(x) 7 log3(x) = 0 => log3(x) = 0 => x = 1

I don't believe that the answer is ln(x)x^(ln(x)-2), since the power rule doesn't apply when you have the variable in the exponent. Do the following instead:y x^ln(x)Taking the natural log of both sides:ln(y)ln(x) * ln(x)ln(y) ln(x)^2Take the derivative of both sides, using the chain rule:1/y * y' 2 ln(x) / xy' 2 ln(x)/ x * yFinally, substitute in the first equation, y x^ln(x):y' 2 ln(x) / x * x^ln(x)y'2 ln(x) * x ^ (ln(x) - 1)Sorry if everything is formatted really badly, this is my first post on answers.com.

int(ln(x2)dx)=xln|x2|-2x int(ln2(x)dx)=x[(ln|x|-2)ln|x|+2]

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