How do you solve ${log}_{10} (4 x) - {log}_{10} (12 + \sqrt{x}) = 2$ $log_10(4x)-log_10(12+sqrtx)=2$ ?

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Answer 1 · 2017-02-05T14:32:10

$x = \frac{125 (10 + \sqrt{73})}{2}$ $x=(125(10+sqrt73))/2$

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

${log}_{10} (4 x) - {log}_{10} (12 + \sqrt{x}) = 2$ $log_10(4x)-log_10(12+sqrtx)=2$ is equivalent to

${log}_{10} (\frac{4 x}{12 + \sqrt{x}}) = 2$ $log_10((4x)/(12+sqrtx))=2$

Hence from definition of log, we have

$(\frac{4 x}{12 + \sqrt{x}}) = 100$ $((4x)/(12+sqrtx))=100$

or $4 x - 100 \sqrt{x} - 1200 = 0$ $4x-100sqrtx-1200=0$ and dividing by $4$ $4$

$x - 25 \sqrt{x} - 300 = 0$ $x-25sqrtx-300=0$ and using quadratic formula

$\sqrt{x} = \frac{25 \pm \sqrt{625 + 1200}}{2} = \frac{25 \pm 5 \sqrt{73}}{2}$ $sqrtx=(25+-sqrt(625+1200))/2=(25+-5sqrt73)/2$

But we cannot have $4 x < 0$ $4x<0$ , hence $\sqrt{x} = \frac{25 + 5 \sqrt{73}}{2}$ $sqrtx=(25+5sqrt73)/2$

$x = \frac{{(25 + 5 \sqrt{73})}^{2}}{4} = \frac{625 + 1825 + 250 \sqrt{73}}{4}$ $x=(25+5sqrt73)^2/4=(625+1825+250sqrt73)/4$

or $\frac{2500 + 250 \sqrt{73}}{4}$ $(2500+250sqrt73)/4$

or $\frac{250 (10 + \sqrt{73})}{4}$ $(250(10+sqrt73))/4$

or $\frac{125 (10 + \sqrt{73})}{2}$ $(125(10+sqrt73))/2$

How do you solve log_10(4x)-log_10(12+sqrtx)=2log10(4x)−log10(12+√x)=2log_10(4x)-log_10(12+sqrtx)=2?