Solve for x ? 2e^2x=4

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Answer 1 · 2018-02-20T04:55:08

#x=ln(2)/2#

#2 e^(2x)=4#

#e^(2x)=2#

Put the natural log function around both sides:

#ln( e^(2x))=ln(2)#

Recall the exponential logarithm property which states that

#log( a^x)=xlog(a)#

and apply it here:

#2xln(e)=ln(2)#

#ln(e)=1# , this is from the basic definition of the natural log function.

#ln(e)= log_e e=1#

since #e^1=e#.

So

#2x=ln(2)#

#x=ln(2)/2#

Answer 2 · 2018-02-20T07:24:46

#x=ln2/2~~0.35#

I'm assuming that your problem is

#2e^(2x)=4#

In this case, we first have to divide by #2# to get

#e^(2x)=2#

If we now take the natural log of both sides, we get

#ln(e^(2x))=ln2#

Let #y=2x#

#<=>ln(e^y)=ln2#

Recall that, #ln(e^y)=y#

#:.y=ln2#

Substituting back #y=2x#, we get

#2x=ln2#

Now, we just have to divide by #2# to isolate #x#:

#x=ln2/2#

Using a calculator, this yields #0.34657...~~0.35#