How do you solve #3-log 7x + 4 = 5#?

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Answer 1 · 2015-12-02T08:34:56

Use the exponential function and a property of logarithms to find that
#x = e^2/7#

We will be using the property that

#e^log(a) = a#

#3-log(7x) + 4 = 5#

#=> -log(7x) = 5 - 3 - 4 = -2#

#=> log(7x) = 2#

#=> e^log(7x) = e^2#

#=> 7x = e^2##" "# (by the above property)

#=> x = e^2/7#