How do you solve ${log}_{x + 2} 1000 = 3$ $log _(x+2) 1000 = 3$ ?

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How do you solve ${log}_{x + 2} 1000 = 3$ $log _(x+2) 1000 = 3$ ?

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Answer 1 · 2016-05-08T16:23:41

$x = 8$ $x =8$ .

Explanation:

Use inversion $a = b^{c}$ $a =b^c$ for $c = {log}_{b} a$ $c=log_b a$ .

Here, $1000 = {(x + 2)}^{3}$ $1000=(x+2)^3$

So, $x + 2 = 1000^{\frac{1}{3}} = {(10^{3})}^{\frac{1}{3}} = 10^{(\frac{1}{3}) (3)} = 10^{1} = 10$ $x+2=1000 ^(1/3)=(10^3)^(1/3)=10^((1/3)(3))=10^1=10$

x=8..

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How do you solve ${log}_{x + 2} 1000 = 3$ $log _(x+2) 1000 = 3$ ?

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