# Thread: Equation with square roots and logs

1. ## Equation with square roots and logs

$\displaystyle 1) \sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}} = log_\frac{1}{2} \left ( x-\frac{25}{4} \right )$ find x

repost your other questions one per thread

2. Help me Please I need the answer
This is from : Singapore Mathematics Olympiad 2001

3. Originally Posted by OnlyMath
$\displaystyle 1) \sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}} = log_\frac{1}{2} \left ( x-\frac{25}{4} \right )$ find x

repost your other questions one per thread

I think, there is no way to solve this monster !!!!!!!!!!!!!!!!!!!!

4. I have the answer. But I want to know how to solve it!

Can someone, a good mathematician, please show me the steps of how to solve it?

Answer: x = 6.750...
(Thanks to Wolfram|Alpha.)
Wolfram alpha drew the graph of:

$\displaystyle \sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}}$ and $\displaystyle log_\frac{1}{2} \left ( x-\frac{25}{4} \right )$ then found where they intercepted to calculate what x was.

Alternate form: $\displaystyle \displaystyle{ln(x-25/4)+\sqrt{x-4 \sqrt{x-4}} ln(4)+\sqrt{4 x-12 \sqrt{x-4}-7} ln(2) = 0}$

I don't think you will get the answer anywhere else unless someone good at maths has a lot of time to solve it.

5. Originally Posted by OnlyMath
$\displaystyle 1) \sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}} = log_\frac{1}{2} \left ( x-\frac{25}{4} \right )$ find x

$\displaystyle \displaystyle\sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}} = log_{0.5} \left(x-\frac{25}{4}\right)$

$\displaystyle \Rightarrow\displaystyle\ 0.5^{\left(\sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}}\right)}=x-6.25$

$\displaystyle L=x-6.25$

$\displaystyle L$ is a line, rising at 45 degrees, crossing the x-axis at x=6.25.

$\displaystyle y=0.5^{\left(\sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}}\right)}$

is positive, so $\displaystyle L$ can only intersect it beyond x=6.25.

$\displaystyle y=0.5^{\left(\sqrt{4x-7-12\sqrt{x-4}} + \sqrt{4x-16\sqrt{x-4}}\right)}=0.5^{\left(\sqrt{4x-16\sqrt{x-4}+(4\sqrt{x-4}-7)}+\sqrt{4x-16\sqrt{4-x}}\right)}$

When x=6.25

$\displaystyle y=0.5^{\left(\sqrt{25-16\sqrt{2.25}+(4\sqrt{2.25}-7)}+\sqrt{25-16\sqrt{2.25}}\right)}$

and since

$\displaystyle 15^2=225\Rightarrow\ (1.5)^2=2.25$

Or...

$\displaystyle \displaystyle\sqrt{2.25}=\sqrt{\frac{9}{4}}=\frac{ 3}{2}$

$\displaystyle x=6.25\Rightarrow\ y=0.5^{\left(\sqrt{18-16(1.5)+4(1.5)}}+\sqrt{25-16(1.5)}\right)}=0.5^{\left(\sqrt{24-24}+\sqrt{25-24}\right)}=0.5^{\left(\sqrt{0}+\sqrt{1}\right)}$

$\displaystyle =0.5$

We can also evaluate the exponent easily when x=8.

$\displaystyle x=8\Rightarrow\ y=0.5^{\left(\sqrt{32-16(2)+4(2)-7}+\sqrt{32-16(2)}\right)}=0.5^{\left(\sqrt{1}+\sqrt{0}\right) }=0.5$

Also, notice that if

$\displaystyle 4\sqrt{x-4}-7=0\Rightarrow\ x-4=\frac{49}{16}\Rightarrow\ x=\frac{105}{16}=6.5625$

$\displaystyle \Rightarrow\ 4x-16\sqrt{x-4}=\frac{49}{4}+16-16\,\left(\frac{7}{4}\right)=0.25$

$\displaystyle x=6.5625\Rightarrow\ y=0.5^{\left(\sqrt{0.25}+\sqrt{0.25}\right)}=0.5^{ \left(0.5+0.5\right)}=0.5$

For what x will $\displaystyle L$ be 0.5 ? and will $\displaystyle y$ be 0.5 there too?

6. Hello, OnlyMath!

I performed a lot of Olympic-level gymnastics,
. . but it got me nowhere . . .

$\displaystyle \text{Solve for }x\!:$

. . $\displaystyle \underbrace{\sqrt{4x-7-12\sqrt{x-4}}}_{A} + \underbrace{\sqrt{4x-16\sqrt{x-4}}}_{B} \;=\; \log_\frac{1}{2}\!\left ( x-\frac{25}{4} \right )$ .[1]

Under radical $\displaystyle A$ we have: .$\displaystyle 4x - 7 - 12\sqrt{x-4}$

Subtract and add 9: .$\displaystyle 4x - 16 - 12\sqrt{x-4} + 9$

. . . . . . . . . . . . .$\displaystyle =\; 4(x-4) - 12\sqrt{x-4} + 9$

. . . . . . . . . . . . .$\displaystyle =\; 4(\sqrt{x-4})^2 - 12\sqrt{x-4} + 9$

. . . . . . . . . . . . .$\displaystyle =\;(2\sqrt{x-4} - 3)^2$

Hence: .$\displaystyle \text{Radical }A \;=\;\sqrt{4x - 7 -12\sqrt{x-4}} \;=\;2\sqrt{x-4} - 3$

Under radical $\displaystyle B$ we have: .$\displaystyle 4x - 16\sqrt{x-4}$

Subtract and add 16: .$\displaystyle 4x-16 - 16\sqrt{x-4} + 16$

. . . . . . . . . . . . . $\displaystyle =\;4\bigg[(x-4) - 4\sqrt{x-4} + 4\bigg]$

. . . . . . . . . . . . . $\displaystyle =\;4\bigg[(\xqrt{x-4})^2 - 4\sqrt{x-4} + 4\bigg]$

. . . . . . . . . . . . . $\displaystyle =\;4\bigg[\sqrt{x-4} - 2\bigg]^2$

Hence: .$\displaystyle \text{Radical }B \;=\;\sqrt{4x-16\sqrt{x-4}} \;=\;2(\sqrt{x-4} - 2)$

Substitute into [1]: .$\displaystyle (2\sqrt{x-4} - 3) + 2(\sqrt{x-4} - 2) \;=\;\log_{\frac{1}{2}}\!\!\left(x - \frac{25}{4}\right)$

And we have: .$\displaystyle 4\sqrt{x-4} - 7 \;=\;\log_{\frac{1}{2}}\!\!\left(x-\frac{25}{4}\right)$

And there is no elementary solution to this transcendental equation.

7. Originally Posted by Soroban
Hello, OnlyMath!

I performed a lot of Olympic-level gymnastics,
. . but it got me nowhere . . .

Under radical $\displaystyle A$ we have: .$\displaystyle 4x - 7 - 12\sqrt{x-4}$

Subtract and add 9: .$\displaystyle 4x - 16 - 12\sqrt{x-4} + 9$

. . . . . . . . . . . . .$\displaystyle =\; 4(x-4) - 12\sqrt{x-4} + 9$

. . . . . . . . . . . . .$\displaystyle =\; 4(\sqrt{x-4})^2 - 12\sqrt{x-4} + 9$

. . . . . . . . . . . . .$\displaystyle =\;(2\sqrt{x-4} - 3)^2$

Hence: .$\displaystyle \text{Radical }A \;=\;\sqrt{4x - 7 -12\sqrt{x-4}} \;=\;2\sqrt{x-4} - 3$

Under radical $\displaystyle B$ we have: .$\displaystyle 4x - 16\sqrt{x-4}$

Subtract and add 16: .$\displaystyle 4x-16 - 16\sqrt{x-4} + 16$

. . . . . . . . . . . . . $\displaystyle =\;4\bigg[(x-4) - 4\sqrt{x-4} + 4\bigg]$

. . . . . . . . . . . . . $\displaystyle =\;4\bigg[(\xqrt{x-4})^2 - 4\sqrt{x-4} + 4\bigg]$

. . . . . . . . . . . . . $\displaystyle =\;4\bigg[\sqrt{x-4} - 2\bigg]^2$

Hence: .$\displaystyle \text{Radical }B \;=\;\sqrt{4x-16\sqrt{x-4}} \;=\;2(\sqrt{x-4} - 2)$

Notice here that for x<8, you will have taken the negative square root of your square on the previous line...

Substitute into [1]: .$\displaystyle (2\sqrt{x-4} - 3) + 2(\sqrt{x-4} - 2) \;=\;\log_{\frac{1}{2}}\!\!\left(x - \frac{25}{4}\right)$

And we have: .$\displaystyle 4\sqrt{x-4} - 7 \;=\;\log_{\frac{1}{2}}\!\!\left(x-\frac{25}{4}\right)$

And there is no elementary solution to this transcendental equation.

Just one error there, Soroban.

For 6.25<x<8...

$\displaystyle 2(2-\sqrt{x-4})$