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 \(\displaystyle a=1995\) 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]

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

Here is my solution:

Spoiler

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)