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How do you solve ${log}_{5} (log 3 x) = 0$ $log_5(log3x) = 0$ ?

1 Answer

Dec 15, 2015

Use the equivalence:
$XXX {log}_{b} (a) = c \Leftrightarrow b^{c} = a$ $color(white)("XXX")color(red)(log_b(a)=c) hArr color(blue)(b^c=a)$
to determine
$XXX x = \frac{10}{3}$ $color(white)("XXX")x=10/3$

Explanation:

${log}_{5} (log (3 x)) = 0$ $log_5(log (3x))=0$

$\Rightarrow 5^{0} = log (3 x)$ $rArr 5^0 = log(3x)$
$\to XXX log (3 x) = 1$ $rarrcolor(white)("XXX")log(3x)=1$

$log (3 x) = 1$ $log(3x)=1$
$\Rightarrow 10^{1} = 3 x XXXXXXXXXXXXX$ $rArr 10^1=3xcolor(white)("XXXXXXXXXXXXX")$ remember the default $log$ $log$ base is $10$ $10$
$\to XXX x = \frac{10}{3}$ $rarrcolor(white)("XXX")x=10/3$

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