How do you solve $log (\frac{1}{100}) = log (10^{x + 2})$ $log (1/100) = log(10^(x+2))$ ?

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How do you solve $log (\frac{1}{100}) = log (10^{x + 2})$ $log (1/100) = log(10^(x+2))$ ?

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Answer 1 · 2018-06-19T09:23:43

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

Explanation:

$using the law of logarithms$ $"using the "color(blue)"law of logarithms"$

$∙ x log x = log y \Leftrightarrow x = y$ $•color(white)(x)logx=logyhArrx=y$

$\frac{1}{100} = 10^{x + 2}$ $1/100=10^(x+2)$

$10^{- 2} = 10^{x + 2}$ $10^(-2)=10^(x+2)$

$since the bases are equal, equate the exponents$ $"since the bases are equal, equate the exponents"$

$x + 2 = - 2 \Rightarrow x = - 4$ $x+2=-2rArrx=-4$

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How do you solve $log (\frac{1}{100}) = log (10^{x + 2})$ $log (1/100) = log(10^(x+2))$ ?

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