log/expoential equations precalculs

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e^x times e^(x+1) = 1

a. combine the exponential expressions to a single expoential expression, 

b. convert to logarthmic eqation.

c. solve equation 

Sep 2nd, 2015

Thank you for the opportunity to help you with your question!

To combine the expression, use the law of exponents,

e^(x)* e^(x+1)=1

Gives, e^(x+x+1)=1                                                                         [Law: (x^m)* (x^n)= x^(m+n) ]

=> e^(2x+1)=1

Now to convert to log, take log of both the sides,

gives, log(e^(2x+1)) = log(1)

Using properties of logarithms,

=> (2x+1)log(e)= 0                                                                          [Laws: log(x^a)= a log(x); log(1) =0 ]

Now, to solve the equation, we have log(e)= 1,

=> 2x+1 = 0

=> x= -(1/2)

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 2nd, 2015

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