Solve an equation involving logarithms

$(\log_{10}x)^2 = 3 \log_{10}x$

Should I do it in this way:

$\log_{10}x^2=3\log_{10}x$

$\frac{\log_{10}x^2}{\log_{10}x}=3$

Is it right?

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Why do you think that $(log_{10} x)^{2} = log_{10} x^2$ – Indrayudh Roy Feb 9 '14 at 18:31
Is it wrong? :( Ooops.... – Kiara Feb 9 '14 at 18:31
We have $log_{10}x^2 = 2log_{10} |x|$ which is not necessarily equal to $(log_{10}x)^{2}$. – Indrayudh Roy Feb 9 '14 at 18:35
Oh...understood...thanks! – Kiara Feb 9 '14 at 18:37

Whatever the number $\log_{10} x$ is lets just call it $y$, so $y = \log_{10} x$. Now $$(\log_{10}x)^2 = 3\log_{10}x$$ becomes $$y^2 = 3y.$$ This equation has two solutions, $y = 0$ and $y = 3$. So your original equation will have two solutions, one is found by solving $$\log_{10}x = 0$$ and the other is found by solving $$\log_{10}x = 3.$$

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Substitute $\log_{10}x=t$ we get $t^2=3t$

$t^2-3t=0$

$t(t-3)=9$

$t=0$

$t=3$

$\log_{10}x=3$

$x=10^{3}$

$\log_{10}x=0$

$x=1$

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