# Logarithmic equation $\log^2_{4}{x}-\log_{2}{x}+4=0$

How to solve this logarithmic equation?

$\log^2_{4}{x}-\log_{2}{x}+4=0$

I do not understand how to start a solution.

• You 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? May 13 '18 at 4:20
• Get rid of bases. To any base, if $log \,x/log \,,2 =u$ then you have a quadratic equation $u^2/4 -u + 4=0$ May 15 '18 at 11:07

$$\log^2_{4}{x}-\log_{2}{x}+4=0$$

$$\log^2_{4}{x}-2\log_{4}{x}+4=0$$

Let $$u=\log_{4}{x}$$

We get $$u^2-2u+4=0$$

Which does not have real solutions.

• HINT: $\log_{a^{c}}{b} = \frac{1}{c}\log_{a}{b} \$ and $\log_{2}{x} = t$
• I strongly advise you to derive the very basic identities of logarithms May 13 '18 at 4:31

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 $x=2^{2+2i\sqrt{3}}$