How do you solve $3 (x - 2) ln 2 + 1 = 100$ $3(x-2)ln2+1=100$ ?

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Answer 1 · 2016-02-07T21:26:03

$x = \frac{33}{ln 2} + 2$ $x=33/ln2+2$

$3 (x - 2) ln 2 + 1 = 100$ $3(x−2)ln2+1=100$

$3 (x - 2) ln 2 = 99$ $3(x−2)ln2=99$

$(x - 2) ln 2 = 33$ $(x−2)ln2=33$

$x - 2 = \frac{33}{ln 2}$ $x−2=33/ln2$

$x = \frac{33}{ln 2} + 2$ $x=33/ln2+2$

How do you solve 3(x−2)ln2+1=1003(x-2)ln2+1=100?