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Answer 1 · 2016-04-05T04:38:19

#x = frac{ln(2) + 2}{5}#

Take the natural logarithm on both sides

#ln(e^{3x} * e^{2x-1}) = ln(2e)#

From the identity

#ln(ab) = ln(a) + ln(b)#

We can simplify the above equation as

#ln(e^{3x}) + ln(e^{2x-1}) = ln(2) + ln(e)#

#3x + (2x-1) = ln(2) + 1#

#5x = ln(2) + 2#

#x = frac{ln(2) + 2}{5}#

Answer 2 · 2016-04-05T08:48:32

A slightly different approach

#=>x = (ln(2)+2)/5#

Given:#" "e^(3x) xx e^(2x-1)=2e#

Compare to #10^2xx10^1 = 10^3 =10^(2+1)#

Using the above method we have;

#" "e^(3x) xx e^(2x-1)=2e" "->" "e^(3x+2x-1)=2e#

Divide both sides by #e#

#e^(3x+2x-2)=2#

#=>e^(5x-2)=2#

Take logs of both sides

#(5x-2)ln(e)=ln(2)#

But #ln(e)=1#

#5x-2=ln(2)#

#=>x=(ln(2)+2)/5#