How do you solve $log (x + 4) = log x + log 4$ $log (x+4)=log x + log 4$ ?

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How do you solve $log (x + 4) = log x + log 4$ $log (x+4)=log x + log 4$ ?

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Answer 1 · 2016-02-07T16:17:19

Put all logs to one side of the equation and solve using the property ${log}_{a} n - {log}_{a} m = {log}_{a} (\frac{n}{m})$ $log_an - log_am = log_a(n/m)$

Explanation:

$log (x + 4) - log x = log 4$ $log(x + 4) - logx = log4$

$log (\frac{\frac{x + 4}{x}}{4}) = 0$ $log(((x + 4)/(x))/4 )= 0$

Convert to exponential form. The base is 10, since nothing is noted in subscript in the log.

$(\frac{\frac{x + 4}{x}}{4}) = 10^{0}$ $(((x + 4)/x) / 4) = 10^0$

$\frac{x + 4}{x} = 1 \times 4$ $(x + 4)/x = 1 xx4$

x + 4 = 4x

4 = 3x

$\frac{4}{3}$ $4/3$ = x

Hopefully this helps.

Answer 2 · 2016-02-07T16:49:10

$x = \frac{4}{3}$ $x=4/3$

Explanation:

$log (x + 4) - log (x) = log (4)$ $log(x+4)-log(x)=log(4)$

$log (\frac{x + 4}{x}) = log (4)$ $log((x+4)/x)=log(4)$

$log (\frac{x + 4}{x}) - log (4) = 0$ $log((x+4)/x)-log(4)=0$

$log (\frac{x + 4}{x} \times \frac{1}{4}) = 0$ $log((x+4)/x xx1/4)=0$

$log (\frac{x + 4}{4 x}) = 0$ $log((x+4)/(4x))=0$

Consider standard form: ${log}_{10} (x) = y \to 10^{y} = x$ $log_10(x)=y -> 10^y=x$

so ${log}_{10} (\frac{x + 4}{4 x}) = 0 \to 10^{0} = \frac{x + 4}{4 x}$ $" " log_10((x+4)/(4x))=0 -> 10^0=(x+4)/(4x)$

Giving: $4 x = x + 4$ $" "4x=x+4$

$3 x = 4$ $3x=4$

$x = \frac{4}{3}$ $x=4/3$

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