Question #c8320

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Question #c8320

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Answer 1 · 2015-01-08T00:28:43

$log (\frac{{0.01}^{2}}{x^{3}}) = - 10.64$ $log(0.01^2/x^3) = -10.64$
I assume that $log$ $log$ means the logarithm with base $10$ $10$ .

By the property $log (\frac{a}{b}) = log (a) - log (b)$ $log(a/b)=log(a)-log(b)$ we get:
$log ({0.01}^{2}) - log (x^{3}) = - 10.64$ $log(0.01^2) - log(x^3) = -10.64$
$log ({(10^{- 2})}^{2}) + 10.64 = log (x^{3})$ $log((10^{-2})^2) +10.64=log(x^3)$
$log (10^{- 4}) + 10.64 = log (x^{3})$ $log(10^{-4}) +10.64=log(x^3)$
By the property $log (a^{b}) = b log (a)$ $log(a^b) =b log(a)$
$- 4 log (10) + 10.64 = 3 log (x)$ $-4 log(10) +10.64=3 log(x)$
$- 4 + 10.64 = 3 log (x)$ $-4 +10.64=3 log(x)$
$6.64 = 3 log (x)$ $6.64=3 log(x)$
$\frac{6.64}{3} = log (x)$ $6.64/3=log(x)$
By definition of logarithm with base $10$ $10$
$x = 10^{\frac{6.64}{3}} \approx 163, 43$ $x=10^{6.64/3} approx 163,43$

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