How do you solve 2 ln x+ ln x^2=32lnx+lnx2=3?

1 Answer
Dec 4, 2015

x = e^(3/4)x=e34

Explanation:

First of all, you need to "unite" the lnln expressions into one.

This can be done with the logarithmic rules:

log_a (n) + log_a (m) = log_a (n * m )loga(n)+loga(m)=loga(nm)

r * log_a(n) = log_a(n^r)rloga(n)=loga(nr)

So, you can transform your equation as follows:

color(white)(xx)2 ln x + ln x^2 = 3×2lnx+lnx2=3

<=> 2 ln x + 2 ln x = 32lnx+2lnx=3

<=> 4 ln x = 34lnx=3

... divide both sides by 44 ...

<=> ln x = 3/4lnx=34

Now, the inverse function for ln x lnx is e^xex which means that both ln(e^x) = xln(ex)=x and e^(ln x)= xelnx=x hold.

This means that you can apply e^xex to both sides of the equation to "get rid" of the logarithm:

<=> e^ln(x) = e^(3/4)eln(x)=e34

<=> x = e^(3/4)x=e34