# 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)