How do you solve $log 2 x = log (x + 16) - 1$ $log2 x = log (x + 16) - 1$ ?

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How do you solve $log 2 x = log (x + 16) - 1$ $log2 x = log (x + 16) - 1$ ?

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Answer 1 · 2016-03-16T01:58:54

$x = \frac{16}{19}$ $x=16/19$

Explanation:

$1$ $1$ . Start by bringing $log (x + 16)$ $log(x+16)$ to the left side of the equation.

$log (2 x) = log (x + 16) - 1$ $log(2x)=log(x+16)-1$

$log (2 x) - log (x + 16) = - 1$ $log(2x)-log(x+16)=-1$

$2$ $2$ . Using the log property, ${log}_{b} (\frac{m}{n}) = {log}_{b} (m) - {log}_{b} (n)$ $log_color(purple)b(color(red)m/color(blue)n)=log_color(purple)b(color(red)m)-log_color(purple)b(color(blue)n)$ , simplify the equation.

$log (\frac{2 x}{x + 16}) = - 1$ $log((2x)/(x+16))=-1$

$3$ $3$ . Convert to exponential form.

$10^{- 1} = (\frac{2 x}{x + 16})$ $10^(-1)=((2x)/(x+16))$

$4$ $4$ Solve for $x$ $x$ .

$\frac{1}{10} = (\frac{2 x}{x + 16})$ $1/10=((2x)/(x+16))$

$\frac{x + 16}{10} = 2 x$ $(x+16)/10=2x$

$x + 16 = 20 x$ $x+16=20x$

$19 x = 16$ $19x=16$

$color(green)(|bar(ul(color(white)(a/a)x=16/19color(white)(a/a)|)))$

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How do you solve $log 2 x = log (x + 16) - 1$ $log2 x = log (x + 16) - 1$ ?

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Explanation:

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How do you solve log2 x = log (x + 16) - 1log2x=log(x+16)−1 log2 x = log (x + 16) - 1?

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How do you solve $log 2 x = log (x + 16) - 1$ $log2 x = log (x + 16) - 1$ ?