What is x if $log (x + 10) - log (3 x - 8) = 2$ $log(x+10) -log (3x-8)= 2$ ?

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Answer 1 · 2018-03-11T22:55:43

see a solution step below..

$log (x + 10) - log (3 x - 8) = 2$ $log(x + 10) - log(3x - 8) = 2$

Recall the law of logarithm..

$log a - log b = \frac{log a}{log b}$ $loga - logb = loga/logb$

Hence;

$\frac{log (x + 10)}{log (3 x - 8)} = 2 \to applying law of logarithm$ $log(x + 10)/log(3x - 8) = 2 -> "applying law of logarithm"$

$log [\frac{x + 10}{3 x - 8}] = 2$ $log[(x + 10)/(3x - 8)] = 2$

For every log, there is a base, and it's default base is $10$ $10$ , since there was no actual base specified in the given question..

Therefore;

${log}_{10} [\frac{x + 10}{3 x - 8}] = 2$ $log_10 [(x + 10)/(3x - 8)] = 2$

Recall again the law of logarithm..

${log}_{a} x = y \Rightarrow x = a^{y}$ $log_a x = y rArr x = a^y$

Hence;

$\frac{x + 10}{3 x - 8} = 10^{2} \to applying law of logarithm$ $(x + 10)/(3x - 8) = 10^2 -> "applying law of logarithm"$

$\frac{x + 10}{3 x - 8} = 100$ $(x + 10)/(3x - 8) = 100$

$\frac{x + 10}{3 x - 8} = \frac{100}{1}$ $(x + 10)/(3x - 8) = 100/1$

$1 (x + 10) = 100 (3 x - 8) \to cross multiplying$ $1(x + 10) = 100(3x - 8) -> "cross multiplying"$

$x + 10 = 300 x - 800$ $x + 10 = 300x - 800$

$300 x - 800 = x + 10 \to rearranging the equation$ $300x - 800 = x + 10 -> "rearranging the equation"$

$300 x - x = 10 + 800 \to collecting like terms$ $300x - x = 10 + 800 ->"collecting like terms"$

$299 x = 810$ $299x = 810$

$x = \frac{810}{299} \to diving both sides by the coefficient of x$ $x = 810/299 -> "diving both sides by the coefficient of x"$

$x = 2.70$ $x = 2.70$

What is x if log(x+10) -log (3x-8)= 2log(x+10)−log(3x−8)=2log(x+10) -log (3x-8)= 2?