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)

$=> x = e^2/7$

How do you solve 3-log 7x + 4 = 53-log 7x + 4 = 5?