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How do I solve for $x$?

$2^{-100x} = (0.5)^{x-4}$

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  • $\begingroup$ is it $(.5)^{x-4}$ or $(.5)^x-4$? $\endgroup$ – Eleven-Eleven Oct 11 '13 at 18:27
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    $\begingroup$ First write $(0.5)^{x-4}=2^{-(x-4)}$. $\endgroup$ – David Mitra Oct 11 '13 at 18:29
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Hint: you need the same base: So $0.5=\frac{1}{2}=2^{-1}$

Thus, $2^{100x}=(2^{-1})^{x-4}=2^{4-x}$

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$$ -100x\ln\left(2\right) = \left(x - 4\right)\ln\left(0.5\right) = x\ln\left(0.5\right)- 4\ln\left(0.5\right)\,, \quad x = {4\ln\left(0.5\right) \over 100\ln\left(2\right) + \ln\left(0.5\right)} $$

$\ln\left(0.5\right) = -\ln\left(2\right)$ $$ \color{#ff0000}{\large x} = {-4\ln\left(2\right) \over 100\ln\left(2\right) - \ln\left(2\right)} = \color{#ff0000}{-\,{4 \over 99}} $$

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