Logarithmic Equation: Solve x^(log_2 x - 6) / 4) = 4/x

[IloveIl]

Please help me solve this problem:

. . . . .\(\displaystyle \large{ x^{\frac{\log_2\, x - 6}{4}}\, =\, \dfrac{4}{x} }\)

Thanks

 

[ksdhart2]

The problem as stated is a bit ambiguous. Did you mean:

\(\displaystyle x^{\dfrac{log_2(x-6)}{4}}=\dfrac{4}{x}\)

or

\(\displaystyle x^{\dfrac{log_2(x)-6}{4}}=\dfrac{4}{x}\)

or something else? In any case, what are your thoughts? What have you tried? I'd note that the exact solution is certain to be awful and messy. Have you considered a numerical approximation method, such as Newton-Raphsonhttp://www.sosmath.com/calculus/diff/der07/der07.html?

 

[lookagain]

The problem as stated is a bit ambiguous. Did you mean:

\(\displaystyle x^{\dfrac{log_2(x-6)}{4}}=\dfrac{4}{x}\)

or

\(\displaystyle x^{\dfrac{log_2(x)-6}{4}}=\dfrac{4}{x}\)

or something else?

The second equation is meant. That is doable.

Suggestion:

Take log to the base 2 of each side.

The exponent on x comes down as a multiplier of a (now) log (base 2) of x.

The right side becomes log(base 2) of 4 - log (base 2) of x.

That right side simplifies to 2 - log (base 2) of x.

You now have a 4 as the denominator on the left-hand side. Multiply both sides of the equation by that 4.

Suggestion: Let a variable, such as y, equal log (base 2) of x, everywhere in the equation.

Now solve that quadratic equation for the values of y and substitute back to find the x-values.