log(x,64)+log(2,x)=1
You asked:
Solve the equation \(\log_{x}\left( 64 \right) + \log_{2}\left( x \right) = 1\) for the variable \(x\).
MathBot Answer:
The 2 solutions to the equation are: \[\begin{aligned}x &= \sqrt{2} \cos{\left(\frac{\sqrt{23} \log{\left(2 \right)}}{2} \right)} + \sqrt{2} i \sin{\left(\frac{\sqrt{23} \log{\left(2 \right)}}{2} \right)} \approx -0.12895561 + 1.4083219 i\\x &= \sqrt{2} \cos{\left(\frac{\sqrt{23} \log{\left(2 \right)}}{2} \right)} - \sqrt{2} i \sin{\left(\frac{\sqrt{23} \log{\left(2 \right)}}{2} \right)} \approx -0.12895561 -1.4083219 i\end{aligned}\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).