# Logarithmic equation's solution [closed]

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, ThomasJun 28 '16 at 13:49

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – colormegone, Claude Leibovici, Watson, user91500, Thomas
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• Is that actually $\ (2.2)^{0.5} \ x \$ , or is it supposed to be $\ 2 \cdot 2^{0.5} \ x \$ ? – colormegone Jun 28 '16 at 3:14
• Could you place parentheses around appropriate parts so as to prevent confusion? – Drew Christensen Jun 28 '16 at 3:17
• It is $2⋅2^{0.5}.x$ – 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

$$(\log_2(x))^2=2\log_2(2\cdot2^{0.5})+2\log_2(x)=3+2\log_2(x)$$

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

$$(\log_2(x))^2-2\log_2(x)-3=0$$

which can be factored as

$$(\log_2(x)+1)(\log_2(x)-3)=0$$

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.