How do you solve #2 ln x+ ln x^2=3#?

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Answer 1 · 2015-12-04T11:19:54

#x = e^(3/4)#

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

This can be done with the logarithmic rules:

#log_a (n) + log_a (m) = log_a (n * m )#

#r * log_a(n) = log_a(n^r)#

So, you can transform your equation as follows:

#color(white)(xx)2 ln x + ln x^2 = 3#

#<=> 2 ln x + 2 ln x = 3#

#<=> 4 ln x = 3#

... divide both sides by #4# ...

#<=> ln x = 3/4#

Now, the inverse function for #ln x # is #e^x# which means that both #ln(e^x) = x# and #e^(ln x)= x# hold.

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

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

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