1. Resolving a logarithmic equation

Hey everyone,

So I've been given the following equation:

and the following interval:

and I have been asked to solve it for a=3 and a=1/3

I assume to do so I must start by using the log(ab)=log(a)+log(b) identity to make it:
$log_{a}((2cos(x)+1)(cos(x)))\geq 0$
but after that I am lost (it is safe to say I suck at logs). Could someone please help me out?

Much appreciated.

2. Re: Resolving a logarithmic equation

$\log_a b=c$ means $a^c = b$. There is not much of a trick. Just take $a$ to the power of each side.

3. Re: Resolving a logarithmic equation

Thanks for the help!
That seems way simpler than I thought this problem would be!
Could you just tell me if I am doing it correctly?

$log_{3}((2cos(x))(cos(x)))\geq 0$

$log_{3}(2cos^{2}(x)+cos(x))\geq 0$

$3^{(2cos^{2}(x)+cos(x))}\geq 0$

$3^{0}\geq (2cos^{2}(x)+cos(x))$

$1\geq (2cos^{2}(x)+cos(x))$

$2cos^{2}(x)+cos(x)-1\leq 0$

$(2cos(x)-1)(cos(x)+1)\leq 0$

$(2cos(x)-1)\leq 0$

$cos(x)\leq 1/2$

OR

$(cos(x)+1)\leq 0$

$cos(x)\leq -1$

And wouldn't that mean that a could equal 3, 1/3, or anything else, and the answer would still be the same?

thanks for the help

Spoiler:
EDIT:
Actually major brain fart on the line where I apply 3^0. Better redo this.

EDIT 2:
Redid the problem. Hope it is right this time:

$log_{3}((2cos(x))(cos(x)))\geq 0$

$log_{3}(2cos^{2}(x)+cos(x))\geq 0$

$3^{(2cos^{2}(x)+cos(x))}\geq 0$

$3\geq (2cos^{2}(x)+cos(x))$

$0\geq 2cos^{2}(x)+cos(x)-3$

$0\geq (2cos(x)-3)(cos(x)+1)$

------------------------------------

$0\geq (2cos(x)-13)$

$\frac{3}{2}\geq cos(x)$

All values of x are negative

OR

$-1\geq cos(x)$

All values of x are positive except x = pi

And since a positive x negative equals it will always be negative (or zero) and therefore all values of x in the given interval resolve the equation

EDIT 3:
Nope, it turns out I was right the first time. Spoilering my whole little... "episode" up there. Well this problem was a roller coaster.

4. Re: Resolving a logarithmic equation

You have:

$\log_a(2\cos^2 x + \cos x)\ge 0$

$\dfrac{\ln(2\cos^2 x + \cos x)}{\ln a} \ge 0$

When $a>1, \ln a >0$, so multiplying both sides by $\ln a$ does not change the inequality. When $a < 1, \ln a < 0$, so multiplying both sides by $\ln a$ flips the inequality (since you are multiplying by a negative number). That is the only difference between when $a=3$ or when $a = \dfrac{1}{3}$