How do you solve $4 log (x - 2) = 7$ $4 log (x-2) = 7$ ?

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Answer 1 · 2018-03-04T06:26:10

$x = 10^{\frac{7}{4}} + 2$ $x=10^(7/4)+2$

${log (x - 2)}^{4} = 7$ $log(x-2)^4=7$

${(x - 2)}^{4} = 10^{7}$ $(x-2)^4=10^7$

$x - 2 = \pm {\sqrt[4]{10}}^{7}$ $x-2=±root(4)10^7$

$x - 2 = \pm 10^{\frac{7}{4}}$ $x-2=±10^(7/4)$

$x = 10^{\frac{7}{4}} + 2$ $x=10^(7/4)+2$

How do you solve 4 log (x-2) = 74log(x−2)=74 log (x-2) = 7?