How do I go about solving this problem: e^2x-1=31?

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Answer 1 · 2018-04-05T08:42:12

#color(blue)(x~~1.732867952)#

I am reading this as:

#e^(2x)-1=31#

#e^(2x)=32#

Taking natural logarithms of both sides:

#2xln(e)=ln(32)#

#ln(e)=1color(white)(888)# ( The logarithm of the base is always 1)

#2x=ln(32)#

#x=ln(32)/2#

#color(blue)(x~~1.732867952)#