How many roots are there to $x=8^{\log_2x} -9^{\log_3x}-4^{\log_2x}+\log_{0.5}{0.25}$?

How many roots are there to $x=8^{\log_2x} -9^{\log_3x}-4^{\log_2x}+\log_{0.5}{0.25}$?

Question

The number of positive values $x$ which satisfy the equation $x=8^{\log_2x} -9^{\log_3x}-4^{\log_2x}+\log_{0.5}{0.25}$ is

1. 0;

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1. 1;

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1. 2;

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Suggestion

Can we evaluate $\log_{0.5}{0.25}$ ?

We know that $8$ is a power of $2$ – could use this fact to rewrite $8^{\log_2x}$?

Solution

We note that $\log_{0.5}0.25=\log_{0.5}(0.5)^2=2$.

If $x>0$ then $8^{\log_2x}=(2^3)^{\log_2x}=(2^{\log_2x})^3=x^3.$

Similarly, $9^{\log_3x}=(3^2)^{\log_3x}=(3^{\log_3x})^2=x^2$ and $4^{\log_2x}=(2^2)^{\log_2x}=(2^{\log_2x})^2=x^2.$

So the equation becomes $x=x^3-x^2-x^2+2,$ that is, $x^3-2x^2-x+2=0.$

We can see that $x=2$ is a solution, so $(x-2)$ is a factor. By inspection, we find $(x-2)(x^2-1)=0$ and so $(x-2)(x-1)(x+1)=0.$

So the roots are $x=-1$, $1$ and $2$.

There are two positive values of $x$ satisfying the equation so the answer is c.