If log5 x - log5(x - 2) = 3, then the value of x, to the nearest hundredth, is?

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Answer 1 · 2017-01-13T00:21:37

$x ≅ 2.02$ $x ~= 2.02$ (Assuming $log 5$ $log5$ represents ${log}_{5}$ $log_5$ )

Assuming $log 5$ $log5$ represents ${log}_{5}$ $log_5$ (log to the base 5), we have:

${log}_{5} x - {log}_{5} (x - 2) = 3$ $log_5x-log_5(x-2)=3$

${log}_{5} (\frac{x}{x - 2}) = 3$ $log_5(x/(x-2)) =3$

$\frac{x}{x - 2} = 5^{3} = 125$ $x/(x-2) = 5^3 = 125$

$x = 125 (x - 2)$ $x=125(x-2)$

$124 x = 250$ $124x = 250$

$x = \frac{250}{124} ≅ 2.02$ $x=250/124 ~= 2.02$ (To the nearest hundredth as requested)

However, to perform a check you should use 8 decimal places:
$x ≅ 2.01612903$ $x~= 2.01612903$