I'm unable to solve this equation further. Could someone have a shot at it and try to solve it and explain it to me please? The equation is $$2\log_{2}(\log_{2}(x))-\log_{2}(\log_{2}2\cdot 2^{0.5}x)=1.$$ I am trying to learn the general rule for solving this kind of log problems. So it does involve a concept and kindly donut flag it. Two people have already given me some good answers.


closed as off-topic by colormegone, Claude Leibovici, Watson, user91500, Thomas Jun 28 '16 at 13:49

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  • $\begingroup$ Is that actually $ \ (2.2)^{0.5} \ x \ $ , or is it supposed to be $ \ 2 \cdot 2^{0.5} \ x \ $ ? $\endgroup$ – colormegone Jun 28 '16 at 3:14
  • 1
    $\begingroup$ Could you place parentheses around appropriate parts so as to prevent confusion? $\endgroup$ – Drew Christensen Jun 28 '16 at 3:17
  • $\begingroup$ It is $2⋅2^{0.5}.x$ $\endgroup$ – Ujjwal Jun 28 '16 at 3:20

Put $\log_2(x)=X$. You have $$\log_2\frac{X^2}{3/2+X}=1=\log_2(2)$$ $$X^2=3+2X\iff X=3\text{ or -1}\iff \log_2 (x)=3\text{ or } \log_2(x)=-1$$ You have to discard $\log_2(x)=-1$ because of the proposed equation.Hence $$\color{red}{x=8}$$


1) Perform algebraic rearrangement

$$2\log_{2}(\log_{2}(x))=1+\log_{2}(\log_{2}(2\cdot 2^{0.5}x))$$

2) Apply laws of logarithms

$$\log_2((\log_2(x))^2)=\log_{2}(2\log_{2}(2\cdot 2^{0.5}x))$$

3) Raise 2 to the power of both sides

$$(\log_2(x))^2=2\log_{2}(2\cdot 2^{0.5}x)$$

4) Apply the laws of logarithms once more


Notice that this can be arranged into a quadratic in $\log_2(x)$, namely


which can be factored as


Therefore, $\log_2(x)=-1$ or $\log_2(x)=3$. This is equivalent to $x=1/2$ or $x=8$. But we aren't yet done! Notice that for 1/2 to be a solution, logarithms have to be defined for negative numbers. This cannot be. Thus, 8 is the only solution.


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