# Thread: The minimum of the function

1. ## The minimum of the function

Hello

The minimum of the function $f(x) = (log_{2}x)^2 + log_{4}x + 1$ is ?

What's the process to solve this problem?

Thanks.

2. Originally Posted by Patrick_John
Hello

The minimum of the function $f(x) = (log_{2}x)^2 + log_{4}x + 1$ is ?

What's the process to solve this problem?

Thanks.
the process is the same as for all problems of this type, find it's derivative and set it equal to zero. if necessary, use the second derivative to verify which of the critical points is a minimum

you will need the change of base formula to change all the logs to ln though

Change of Base Formula for Logarithms:

$\log_a b = \frac { \log_c b}{ \log_c a}$

or maybe noticing that $\log_4 x = \log_2 \sqrt {x}$ will simplify the problem a bit. but i'd go with my first suggestion

3. Originally Posted by Patrick_John
Hello

The minimum of the function $f(x) = (log_{2}x)^2 + log_{4}x + 1$ is ?

What's the process to solve this problem?

Thanks.
Do what Jhevon did,
$\log_2^2 x + \log_2 \sqrt{x} + 1$

Rewrite as,
$\log_2^2 x + \frac{1}{2}\log_2 x+1$

Let $y=\log_2^2 x$ to get,

$y^2 + \frac{1}{2}y+1$

This is a parabola, you can use the formula to obtain its minimum.

4. Originally Posted by ThePerfectHacker
Do what Jhevon did,
$\log_2^2 x + \log_2 \sqrt{x} + 1$

Rewrite as,
$\log_2^2 x + \frac{1}{2}\log_2 x+1$

Let $y=\log_2^2 x$ to get,

$y^2 + \frac{1}{2}y+1$

This is a parabola, you can use the formula to obtain its minimum.
Thanks for that TPH, I forgot Patrick isn't in Calculus. I see "minimum" and i switched to calc mode. doing it using calc is fun though, the precalc method was so anticlimactic

5. Thanks for the help.

But since I'm not very aware of what is a minimum, I don't know how to get it using the formula, could somebody explain in more detail how to do it?

Also, why did you rewrite $\log_2 \sqrt{x}$ as $\frac{1}{2}\log_2 x$ ?

6. Hello, Patrick_John!

I will assume that we are not allowed to use Calculus . . .

Find the minimum of the function: . $f(x) \:= \:(\log_{2}x)^2 + \log_{4}x + 1$
The logs have two different bases; we'll make them the same.

Let $\log_4x = P$

. . Then: . $4^P = x\quad\Rightarrow\quad (2^2)^P = x\quad\Rightarrow\quad 2^{2P} = x$

. . Take logs (base 2): . $\log_2\left(2^{2P}\right) = \log_2x\quad\Rightarrow\quad2P\!\cdot\!\log_22 = \log_2x$

. . Then: . $2P = \log_2x\quad\Rightarrow\quad P = \frac{1}{2}\log_2x$
. . Hence: . $\log_4x = \frac{1}{2}\log_2x$

Substitute into the original equation: . $f(x)\:=\:\left(\log_2x\right)^2 + \frac{1}{2}\log_2x + 1$

Let $z = \log_2x$
Then we have: . $f(z) \:=\:z^2 + \frac{1}{2}z + 1$

This is an up-opening parabola; its minimum is at its vertex.

The vertex formula is: . $\frac{\text{-}b}{2a}$
We have: . $a = 1,\:b = \frac{1}{2}$
Hence, the vertex is at: . $z \:=\:\frac{\text{-}\frac{1}{2}}{2(1)} \:=\:-\frac{1}{4}$

Therefore, the minimum is: . $f\left(\text{-}\frac{1}{4}\right)\;=\;\left(\text{-}\frac{1}{4}\right)^2 + \frac{1}{2}\left(\text{-}\frac{1}{4}\right) + 1 \;=\;\boxed{\frac{15}{16}}$