How to solve this logarithmic equation?
$\log^2_{4}{x}-\log_{2}{x}+4=0$
I do not understand how to start a solution.
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2You have logs to different bases. First, you should change the base of one of them so that you have logs all to the same base. Then what should you do? – Eric Towers May 13 '18 at 04:20
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Get rid of bases. To any base, if $ log ,x/log ,,2 =u$ then you have a quadratic equation $ u^2/4 -u + 4=0$ – Narasimham May 15 '18 at 11:07
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- HINT: $ \log_{a^{c}}{b} = \frac{1}{c}\log_{a}{b} \ $ and $\log_{2}{x} = t$
Vladislav Kharlamov
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I strongly advise you to derive the very basic identities of logarithms – Vladislav Kharlamov May 13 '18 at 04:31
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$$\log^2_{4}{x}-\log_{2}{x}+4=0$$
$$\log^2_{4}{x}-2\log_{4}{x}+4=0$$
Let $$u=\log_{4}{x}$$
We get $$u^2-2u+4=0$$
Which does not have real solutions.
Mohammad Riazi-Kermani
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The expression is equivalent to $\frac{log^2(x)}{log^2(4)}-\frac{log^2(x)}{log^2(2)}+4=0$ but like you've been told, does not have real solutions. Actually the complex solutions are $x=2^{2-2i\sqrt{3}}$ and $x=2^{2+2i\sqrt{3}}$
Karen
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