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

[anemone]

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

 

[MarkFL]

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

Here is my solution:

Spoiler

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

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

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

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

Applying the log property:

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

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

Applying the log property:

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

we obtain:

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

Applying the log property:

$$\log_a(a)=1$$

we have:

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

Multiply through by 2:

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

Writing in standard quadratic form, there results:

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

Applying the quadratic formula, we find:

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

Hence:

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

The product $p$ of these positive roots is:

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

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

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

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

 

[anemone]

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

Here is my solution:

Spoiler

Let $$a=1995$$ and the equation may be written: $$a^{\frac{1}{2}}x^{\log_a(x)}=x^2$$ Taking the log of base $a$ of both sides, we obtain: $$\log_a\left(a^{\frac{1}{2}}x^{\log_a(x)} \right)=\log_a\left(x^2 \right)$$ Applying the log property: $$\log_a(bc)=\log_a(b)+\log_a(c)$$ on the left side, we have: $$\log_a\left(a^{\frac{1}{2}} \right)+\log_a\left(x^{\log_a(x)} \right)=\log_a\left(x^2 \right)$$ Applying the log property: $$\log_a\left(b^c \right)=c\log_a(b)$$ we obtain: $$\frac{1}{2}\log_a\left(a \right)+\log_a(x)\log_a\left(x \right)=2\log_a\left(x \right)$$ Applying the log property: $$\log_a(a)=1$$ we have: $$\frac{1}{2}+\log_a^2(x)=2\log_a\left(x \right)$$ Multiply through by 2: $$1+2\log_a^2(x)=4\log_a\left(x \right)$$ Writing in standard quadratic form, there results: $$2\log_a^2(x)-4\log_a\left(x \right)+1=0$$ Applying the quadratic formula, we find: $$\log_a(x)=1\pm\frac{1}{\sqrt{2}}$$ Hence: $$x=a^{1\pm\frac{1}{\sqrt{2}}}$$ The product $p$ of these positive roots is: $$p=a^{1+\frac{1}{\sqrt{2}}}a^{1-\frac{1}{\sqrt{2}}}=a^2$$ Using the given value $a=1995$, we find: $$p=1995^2=(2000-5)^2=2000^2-2\cdot2000\cdot5+5^2=1000k+25$$ 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)