How do you solve $4 {log}_{5} (x + 1) = 4.8$ $4log_5(x+1)=4.8$ ?

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Answer 1 · 2016-08-16T02:02:57

$x ≅ 5.8986$ $x~=5.8986$

$4 {log}_{5} (x + 1) = 4.8$ $4log_5(x+1)=4.8$

${log}_{5} (x + 1) = \frac{4.8}{4} = 1.2$ $log_5(x+1) = 4.8/4 = 1.2$

$x + 1 = 5^{1.2} ≅ 6.8986$ $x+1 = 5^1.2 ~= 6.8986$

$x = ≅ 6.8986 - 1 ≅ 5.8986$ $x=~= 6.8986-1 ~= 5.8986$

How do you solve 4log5(x+1)=4.84log_5(x+1)=4.8?