How do you solve $2 log (x + 7) = log 9$ $2 Log (x+7) = Log 9$ ?

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How do you solve $2 log (x + 7) = log 9$ $2 Log (x+7) = Log 9$ ?

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Answer 1 · 2015-12-03T11:59:40

$x = - 4$ $x=-4$

Explanation:

$2 \cdot log (x + 7) = log 9$ $2*log(x+7)=log9$
$\Rightarrow \frac{1}{2} \cdot 2 \cdot log (x + 7) = \frac{1}{2} \cdot log 9$ $=>1/2*2*log(x+7)=1/2*log9$
$\Rightarrow log (x + 7) = log (9^{\frac{1}{2}})$ $=>log(x+7)=log(9^(1/2))$ (the logarithm of a second root is the logarithm of the number divided by 2)
$\Rightarrow log (x + 7) = log 3$ $=>log(x+7)=log3$
$\Rightarrow x + 7 = 10^{log 3}$ $=>x+7=10^log3$
Add -7 to both sides:
$x + 7 - 7 = 3 - 7$ $x+7-7=3-7$
$\Rightarrow x = - 4$ $=>x=-4$

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How do you solve $2 log (x + 7) = log 9$ $2 Log (x+7) = Log 9$ ?

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