What are the last three digits of the product of the positive roots?

anemone

MHB POTW Director
Staff member
What is the last three digits of the product of the positive roots of $\large\sqrt{1995}x^{\log_{1995} x}=x^2$.

MarkFL

Staff member
Re: What is the last three digits of the product of the positive roots?

Here is my solution:

Let $$\displaystyle a=1995$$ and the equation may be written:

$$\displaystyle a^{\frac{1}{2}}x^{\log_a(x)}=x^2$$

Taking the log of base $a$ of both sides, we obtain:

$$\displaystyle \log_a\left(a^{\frac{1}{2}}x^{\log_a(x)} \right)=\log_a\left(x^2 \right)$$

Applying the log property:

$$\displaystyle \log_a(bc)=\log_a(b)+\log_a(c)$$ on the left side, we have:

$$\displaystyle \log_a\left(a^{\frac{1}{2}} \right)+\log_a\left(x^{\log_a(x)} \right)=\log_a\left(x^2 \right)$$

Applying the log property:

$$\displaystyle \log_a\left(b^c \right)=c\log_a(b)$$

we obtain:

$$\displaystyle \frac{1}{2}\log_a\left(a \right)+\log_a(x)\log_a\left(x \right)=2\log_a\left(x \right)$$

Applying the log property:

$$\displaystyle \log_a(a)=1$$

we have:

$$\displaystyle \frac{1}{2}+\log_a^2(x)=2\log_a\left(x \right)$$

Multiply through by 2:

$$\displaystyle 1+2\log_a^2(x)=4\log_a\left(x \right)$$

Writing in standard quadratic form, there results:

$$\displaystyle 2\log_a^2(x)-4\log_a\left(x \right)+1=0$$

Applying the quadratic formula, we find:

$$\displaystyle \log_a(x)=1\pm\frac{1}{\sqrt{2}}$$

Hence:

$$\displaystyle x=a^{1\pm\frac{1}{\sqrt{2}}}$$

The product $p$ of these positive roots is:

$$\displaystyle p=a^{1+\frac{1}{\sqrt{2}}}a^{1-\frac{1}{\sqrt{2}}}=a^2$$

Using the given value $a=1995$, we find:

$$\displaystyle p=1995^2=(2000-5)^2=2000^2-2\cdot2000\cdot5+5^2=1000k+25$$

Thus the last 3 digits of the product of the roots is $025$.

anemone

MHB POTW Director
Staff member
Re: What is the last three digits of the product of the positive roots?

Here is my solution:
Let $$\displaystyle a=1995$$ and the equation may be written: $$\displaystyle a^{\frac{1}{2}}x^{\log_a(x)}=x^2$$ Taking the log of base $a$ of both sides, we obtain: $$\displaystyle \log_a\left(a^{\frac{1}{2}}x^{\log_a(x)} \right)=\log_a\left(x^2 \right)$$ Applying the log property: $$\displaystyle \log_a(bc)=\log_a(b)+\log_a(c)$$ on the left side, we have: $$\displaystyle \log_a\left(a^{\frac{1}{2}} \right)+\log_a\left(x^{\log_a(x)} \right)=\log_a\left(x^2 \right)$$ Applying the log property: $$\displaystyle \log_a\left(b^c \right)=c\log_a(b)$$ we obtain: $$\displaystyle \frac{1}{2}\log_a\left(a \right)+\log_a(x)\log_a\left(x \right)=2\log_a\left(x \right)$$ Applying the log property: $$\displaystyle \log_a(a)=1$$ we have: $$\displaystyle \frac{1}{2}+\log_a^2(x)=2\log_a\left(x \right)$$ Multiply through by 2: $$\displaystyle 1+2\log_a^2(x)=4\log_a\left(x \right)$$ Writing in standard quadratic form, there results: $$\displaystyle 2\log_a^2(x)-4\log_a\left(x \right)+1=0$$ Applying the quadratic formula, we find: $$\displaystyle \log_a(x)=1\pm\frac{1}{\sqrt{2}}$$ Hence: $$\displaystyle x=a^{1\pm\frac{1}{\sqrt{2}}}$$ The product $p$ of these positive roots is: $$\displaystyle p=a^{1+\frac{1}{\sqrt{2}}}a^{1-\frac{1}{\sqrt{2}}}=a^2$$ Using the given value $a=1995$, we find: $$\displaystyle p=1995^2=(2000-5)^2=2000^2-2\cdot2000\cdot5+5^2=1000k+25$$ Thus the last 3 digits of the product of the roots is $025$.
Aww...it's amazing that you solved it so quickly and posted it with your well explained steps, I want to say thank you for participating, MarkFL!(heart)