What is the solution for 1/2=e^(-(ln2)t)?

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Feb 9, 2018

Answer:

# \ #

# \mbox{The solution is:} \qquad \qquad \quad t \ = \ 1. #

Explanation:

# \ #

# \mbox{We can proceed as follows:} #

# \qquad \qquad e^{ -(ln2) t } \ = \ 1/2 #

# \qquad \qquad [ e^(ln2) ]^{ - t } \ = \ 1/2 \qquad \qquad \qquad \qquad \mbox{multiplication rule for exponents} #

# \qquad \qquad \qquad \ \ [ 2 ]^{ - t } \ = \ 1/2 \qquad \qquad \qquad \ \ \ e^x \quad \mbox{and} \quad ln(x) \ \ \mbox{are inverses} #

# \qquad \qquad \qquad \qquad 2^{ - t } \ = \ 1/2 #

# \qquad \qquad \qquad \qquad 2^{ - t } \ = \ 2^{-1} #

# \qquad :. \qquad \quad - t \ = \ -1 #

# \qquad :. \qquad \qquad \quad \ t \ = \ 1. #

# \ #

# \mbox{So the solution is:} #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad t \ = \ 1. #

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